{
  "cells": [
    {
      "cell_type": "markdown",
      "metadata": {},
      "source": [
        "### News"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {},
      "source": [
        "Placeholder"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {},
      "source": [
        "### Installation"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {},
      "outputs": [],
      "source": [
        "# Placeholder"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": null,
      "metadata": {},
      "outputs": [],
      "source": [
        "# Placeholder"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "VIy3QkjW1O4R"
      },
      "source": [
        "### Unsloth"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "-B3HIT0t6nc0"
      },
      "source": [
        "We're also introducing how you can do `GSPO` inside of Unsloth as well!\n",
        "\n",
        "The goal of this notebook is to make a vision language model solve maths problems via reinforcement learning given an image input like below:\n",
        "\n",
        "<img src=\"https://raw.githubusercontent.com/lupantech/MathVista/main/assets/our_new_3_datasets.png\" alt=\"Alt text\" height=\"256\">"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "0cI_m3OJWja8"
      },
      "source": []
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        },
        "id": "DkIvEkIIkEyB",
        "outputId": "76271524-66e5-45d4-9ae8-89268a9f4bbe"
      },
      "outputs": [
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "🦥 Unsloth: Will patch your computer to enable 2x faster free finetuning.\n"
          ]
        },
        {
          "name": "stderr",
          "output_type": "stream",
          "text": [
            "/usr/local/lib/python3.12/dist-packages/pydantic/_internal/_generate_schema.py:2249: UnsupportedFieldAttributeWarning: The 'repr' attribute with value False was provided to the `Field()` function, which has no effect in the context it was used. 'repr' is field-specific metadata, and can only be attached to a model field using `Annotated` metadata or by assignment. This may have happened because an `Annotated` type alias using the `type` statement was used, or if the `Field()` function was attached to a single member of a union type.\n",
            "  warnings.warn(\n",
            "/usr/local/lib/python3.12/dist-packages/pydantic/_internal/_generate_schema.py:2249: UnsupportedFieldAttributeWarning: The 'frozen' attribute with value True was provided to the `Field()` function, which has no effect in the context it was used. 'frozen' is field-specific metadata, and can only be attached to a model field using `Annotated` metadata or by assignment. This may have happened because an `Annotated` type alias using the `type` statement was used, or if the `Field()` function was attached to a single member of a union type.\n",
            "  warnings.warn(\n"
          ]
        },
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "INFO 10-14 12:53:27 [__init__.py:244] Automatically detected platform cuda.\n",
            "ERROR 10-14 12:53:29 [fa_utils.py:57] Cannot use FA version 2 is not supported due to FA2 is only supported on devices with compute capability >= 8\n",
            "🦥 Unsloth Zoo will now patch everything to make training faster!\n",
            "==((====))==  Unsloth 2025.10.2: Fast Qwen3_Vl patching. Transformers: 4.57.0. vLLM: 0.9.2.\n",
            "   \\\\   /|    Tesla T4. Num GPUs = 1. Max memory: 14.741 GB. Platform: Linux.\n",
            "O^O/ \\_/ \\    Torch: 2.7.0+cu126. CUDA: 7.5. CUDA Toolkit: 12.6. Triton: 3.2.0\n",
            "\\        /    Bfloat16 = FALSE. FA [Xformers = 0.0.30. FA2 = False]\n",
            " \"-____-\"     Free license: http://github.com/unslothai/unsloth\n",
            "Unsloth: Fast downloading is enabled - ignore downloading bars which are red colored!\n"
          ]
        },
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        {
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      ],
      "source": [
        "from unsloth import FastVisionModel\n",
        "import torch\n",
        "max_seq_length = 16384 # Must be this long for VLMs\n",
        "lora_rank = 16 # Larger rank = smarter, but slower\n",
        "\n",
        "model, tokenizer = FastVisionModel.from_pretrained(\n",
        "    model_name = \"unsloth/Qwen3-VL-8B-Instruct-unsloth-bnb-4bit\",\n",
        "    max_seq_length = max_seq_length,\n",
        "    load_in_4bit = True, # False for LoRA 16bit\n",
        "    fast_inference = False, # Enable vLLM fast inference\n",
        "    gpu_memory_utilization = 0.8, # Reduce if out of memory\n",
        "\n",
        ")"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "aOXrl8iLQx6S"
      },
      "source": [
        "In Unsloth, we share vLLM's weights directly, reducing VRAM usage by > 50%. vLLM also does not yet support LoRA on the vision layers, so we can only add them on the language layers. Vision GRPO still works though!"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 4,
      "metadata": {
        "id": "pmZ6zZ5AQu7I"
      },
      "outputs": [],
      "source": [
        "model = FastVisionModel.get_peft_model(\n",
        "    model,\n",
        "    finetune_vision_layers     = False, # False if not finetuning vision layers\n",
        "    finetune_language_layers   = True,  # False if not finetuning language layers\n",
        "    finetune_attention_modules = True,  # False if not finetuning attention layers\n",
        "    finetune_mlp_modules       = True,  # False if not finetuning MLP layers\n",
        "\n",
        "    r = 16,           # The larger, the higher the accuracy, but might overfit\n",
        "    lora_alpha = 16,  # Recommended alpha == r at least\n",
        "    lora_dropout = 0,\n",
        "    bias = \"none\",\n",
        "    random_state = 3407,\n",
        "    use_rslora = False,  # We support rank stabilized LoRA\n",
        "    loftq_config = None, # And LoftQ\n",
        "    use_gradient_checkpointing = \"unsloth\", # Reduces memory usage\n",
        "    # target_modules = \"all-linear\", # Optional now! Can specify a list if needed\n",
        ")"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "7KGgPgk_5S8r"
      },
      "source": [
        "### Data Prep\n",
        "<a name=\"Data\"></a>\n",
        "\n",
        "`AI4Math/MathVista` is a dataset that involves using images to solve logic and math problems.\n",
        "\n",
        "For this notebook, we will only use math problems with numeric answers for simpilicity."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 5,
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        },
        "id": "7zM1VPx5KcpF",
        "outputId": "b86fe892-2df6-43c5-c6cf-48ac5f94ab84"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "d8f808d13e954c8e858d4ea65c1c0e0f",
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            "text/plain": [
              "README.md: 0.00B [00:00, ?B/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        },
        {
          "data": {
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              "model_id": "03ad9a740fa6421ba12a816c3d5ab56f",
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            },
            "text/plain": [
              "data/testmini-00000-of-00001-725687bf7a1(…):   0%|          | 0.00/142M [00:00<?, ?B/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        },
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        {
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            "text/plain": [
              "data/test-00001-of-00002-6a611c71596db30(…):   0%|          | 0.00/386M [00:00<?, ?B/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        },
        {
          "data": {
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              "model_id": "c034720c776d420c89402a03b39665bd",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "Generating testmini split:   0%|          | 0/1000 [00:00<?, ? examples/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        },
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "a7984eb31e834d3c831f04c3d3ba53e5",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "Generating test split:   0%|          | 0/5141 [00:00<?, ? examples/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        }
      ],
      "source": [
        "from datasets import load_dataset\n",
        "from trl import GRPOConfig, GRPOTrainer\n",
        "\n",
        "dataset = load_dataset(\"AI4Math/MathVista\", split = \"testmini\")"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "r0CeDQrm6BWW"
      },
      "source": [
        "We filter the dataset to keep only float or numeric answers:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 6,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 49,
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        },
        "id": "iw8EVJmp5rsC",
        "outputId": "cbb80ec8-aa3c-4f94-e4e8-0ac554ab2d4b"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "cd3d4231afa64f20805026d750be7d4e",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "Filter:   0%|          | 0/1000 [00:00<?, ? examples/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        }
      ],
      "source": [
        "def is_numeric_answer(example):\n",
        "    try:\n",
        "        float(example[\"answer\"])\n",
        "        return True\n",
        "    except:\n",
        "        return False\n",
        "\n",
        "dataset = dataset.filter(is_numeric_answer)"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "IAPtJzy_5uLh"
      },
      "source": [
        "We also resize the images to be 512 by 512 pixels to make the images managable in context length. We also convert them to RGB so they are compatible for training!"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 7,
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        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 81,
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        },
        "id": "tT8WNP1A5tsh",
        "outputId": "2492c6eb-0a9d-42f0-886b-1fe9051b3376"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "c00ef2e7b0db41d5870b651c76851b5b",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "Map:   0%|          | 0/566 [00:00<?, ? examples/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        },
        {
          "data": {
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              "model_id": "5312c7f21f4c4852b6a785f8539746f5",
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              "Map:   0%|          | 0/566 [00:00<?, ? examples/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        }
      ],
      "source": [
        "# Resize to (512, 512)\n",
        "def resize_images(example):\n",
        "    image = example[\"decoded_image\"]\n",
        "    image = image.resize((512, 512))\n",
        "    example[\"decoded_image\"] = image\n",
        "    return example\n",
        "dataset = dataset.map(resize_images)\n",
        "\n",
        "# Then convert to RGB\n",
        "def convert_to_rgb(example):\n",
        "    image = example[\"decoded_image\"]\n",
        "    if image.mode != \"RGB\":\n",
        "        image = image.convert(\"RGB\")\n",
        "    example[\"decoded_image\"] = image\n",
        "    return example\n",
        "dataset = dataset.map(convert_to_rgb)"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "D2WpGKjZ7mHI"
      },
      "source": [
        "We then create the conversational template that is needed to collate the dataset for RL:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 8,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 49,
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          ]
        },
        "id": "-lvgcXGjk_a6",
        "outputId": "cc2ee2bb-8113-4c8c-adeb-4304caeaf746"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "bf72683ee80e41c7a2f8c5fdec264286",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "Map:   0%|          | 0/566 [00:00<?, ? examples/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        }
      ],
      "source": [
        "# Define the delimiter variables for clarity and easy modification\n",
        "REASONING_START = \"<REASONING>\"\n",
        "REASONING_END = \"</REASONING>\"\n",
        "SOLUTION_START = \"<SOLUTION>\"\n",
        "SOLUTION_END = \"</SOLUTION>\"\n",
        "\n",
        "def make_conversation(example):\n",
        "    # Define placeholder constants if they are not defined globally\n",
        "    # The user's text prompt\n",
        "    text_content = (\n",
        "        f\"{example['question']}. Also first provide your reasoning or working out\"\\\n",
        "        f\" on how you would go about solving the question between {REASONING_START} and {REASONING_END}\"\n",
        "        f\" and then your final answer between {SOLUTION_START} and (put a single float here) {SOLUTION_END}\"\n",
        "    )\n",
        "\n",
        "    # Construct the prompt in the desired multi-modal format\n",
        "    prompt = [\n",
        "        {\n",
        "            \"role\": \"user\",\n",
        "            \"content\": [\n",
        "                {\"type\": \"image\"},  # Placeholder for the image\n",
        "                {\"type\": \"text\", \"text\": text_content},  # The text part of the prompt\n",
        "            ],\n",
        "        },\n",
        "    ]\n",
        "    # The actual image data is kept separate for the processor\n",
        "    return {\"prompt\": prompt, \"image\": example[\"decoded_image\"], \"answer\": example[\"answer\"]}\n",
        "\n",
        "train_dataset = dataset.map(make_conversation)\n",
        "\n",
        "# We're reformatting dataset like this because decoded_images are the actual images\n",
        "# The \"image\": example[\"decoded_image\"] does not properly format the dataset correctly\n",
        "\n",
        "# 1. Remove the original 'image' column\n",
        "train_dataset = train_dataset.remove_columns(\"image\")\n",
        "\n",
        "# 2. Rename 'decoded_image' to 'image'\n",
        "train_dataset = train_dataset.rename_column(\"decoded_image\", \"image\")"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "uOz-lAoI5fLW"
      },
      "source": [
        "Now let's apply the chat template across the entire dataset:"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 9,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 49,
          "referenced_widgets": [
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        },
        "id": "ZaxwJAqS5d1L",
        "outputId": "12b3a58d-7265-4254-e343-aa0a4f11a1e0"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.jupyter.widget-view+json": {
              "model_id": "2b6d71e2012f4153a105ffbeeb867192",
              "version_major": 2,
              "version_minor": 0
            },
            "text/plain": [
              "Map:   0%|          | 0/566 [00:00<?, ? examples/s]"
            ]
          },
          "metadata": {},
          "output_type": "display_data"
        }
      ],
      "source": [
        "train_dataset = train_dataset.map(\n",
        "    lambda example: {\n",
        "        \"prompt\": tokenizer.apply_chat_template(\n",
        "            example[\"prompt\"],\n",
        "            tokenize = False,\n",
        "            add_generation_prompt = True, # Must add assistant\n",
        "        )\n",
        "    }\n",
        ")"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "VEs2HiThleic"
      },
      "source": [
        "## Reward functions\n",
        "\n",
        "We now define some basic formatting rewards functions to see if reasoning starts and ends, and also another to see if the answers were written correctly.\n",
        "\n",
        "We also try to fix the `addCriterion` issue as described in our [blog post](https://docs.unsloth.ai/new/vision-reinforcement-learning-vlm-rl#qwen-2.5-vl-vision-rl-issues-and-quirks)"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 10,
      "metadata": {
        "id": "cXk993X6C2ZZ"
      },
      "outputs": [],
      "source": [
        "# Reward functions\n",
        "import re\n",
        "\n",
        "def formatting_reward_func(completions,**kwargs):\n",
        "    import re\n",
        "    thinking_pattern = f'{REASONING_START}(.*?){REASONING_END}'\n",
        "    answer_pattern = f'{SOLUTION_START}(.*?){SOLUTION_END}'\n",
        "\n",
        "    scores = []\n",
        "    for completion in completions:\n",
        "        score = 0\n",
        "        thinking_matches = re.findall(thinking_pattern, completion, re.DOTALL)\n",
        "        answer_matches = re.findall(answer_pattern, completion, re.DOTALL)\n",
        "        if len(thinking_matches) == 1:\n",
        "            score += 1.0\n",
        "        if len(answer_matches) == 1:\n",
        "            score += 1.0\n",
        "\n",
        "        # Fix up addCriterion issues\n",
        "        # See https://docs.unsloth.ai/new/vision-reinforcement-learning-vlm-rl#qwen-2.5-vl-vision-rl-issues-and-quirks\n",
        "        # Penalize on excessive addCriterion and newlines\n",
        "        if len(completion) != 0:\n",
        "            removal = completion.replace(\"addCriterion\", \"\").replace(\"\\n\", \"\")\n",
        "            if (len(completion)-len(removal))/len(completion) >= 0.5:\n",
        "                score -= 2.0\n",
        "\n",
        "        scores.append(score)\n",
        "    return scores\n",
        "\n",
        "\n",
        "def correctness_reward_func(prompts, completions, answer, **kwargs) -> list[float]:\n",
        "    answer_pattern = f'{SOLUTION_START}(.*?){SOLUTION_END}'\n",
        "\n",
        "    responses = [re.findall(answer_pattern, completion, re.DOTALL) for completion in completions]\n",
        "    q = prompts[0]\n",
        "    print('-'*20, f\"Question:\\n{q}\", f\"\\nAnswer:\\n{answer[0]}\", f\"\\nResponse:{completions[0]}\")\n",
        "    return [\n",
        "        2.0 if len(r)==1 and a == r[0].replace('\\n','') else 0.0\n",
        "        for r, a in zip(responses, answer)\n",
        "    ]"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "lrst4oDS5AFO"
      },
      "source": [
        "Here is the first example prompt in the dataset"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 11,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 105
        },
        "id": "nXjAu-nQwpuI",
        "outputId": "7edd6d1c-88fc-4d49-dbbf-b903ccb4edfd"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.google.colaboratory.intrinsic+json": {
              "type": "string"
            },
            "text/plain": [
              "\"<|im_start|>user\\n<|vision_start|><|image_pad|><|vision_end|>When a spring does work on an object, we cannot find the work by simply multiplying the spring force by the object's displacement. The reason is that there is no one value for the force-it changes. However, we can split the displacement up into an infinite number of tiny parts and then approximate the force in each as being constant. Integration sums the work done in all those parts. Here we use the generic result of the integration.\\r\\n\\r\\nIn Figure, a cumin canister of mass $m=0.40 \\\\mathrm{~kg}$ slides across a horizontal frictionless counter with speed $v=0.50 \\\\mathrm{~m} / \\\\mathrm{s}$. It then runs into and compresses a spring of spring constant $k=750 \\\\mathrm{~N} / \\\\mathrm{m}$. When the canister is momentarily stopped by the spring, by what distance $d$ is the spring compressed?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\\n<|im_start|>assistant\\n\""
            ]
          },
          "execution_count": 11,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "train_dataset[0][\"prompt\"]"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 12,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 70
        },
        "id": "oKBOl4K1wpWx",
        "outputId": "9826ce80-92c0-4c8a-a35e-60945976f3de"
      },
      "outputs": [
        {
          "data": {
            "application/vnd.google.colaboratory.intrinsic+json": {
              "type": "string"
            },
            "text/plain": [
              "'<|im_start|>user\\n<|vision_start|><|image_pad|><|vision_end|>Move the ruler to measure the length of the nail to the nearest inch. The nail is about (_) inches long.. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\\n<|im_start|>assistant\\n'"
            ]
          },
          "execution_count": 12,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "train_dataset[100][\"prompt\"]"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "8YpenSUIAczo"
      },
      "source": [
        "<a name=\"Inference\"></a>\n",
        "### Inference\n",
        "Now let's try the model on the hundredth sample of the train dataset without training.\n"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 13,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "xGa0rueD5Mfh",
        "outputId": "b3cb5ab7-8ab1-44cd-a1f3-b3b7daf1eb34"
      },
      "outputs": [
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "<REASONING>\n",
            "To measure the length of the nail to the nearest inch, I need to determine how long the nail is by comparing it to the ruler.\n",
            "\n",
            "Step 1: Identify the starting point of the nail.\n",
            "The head of the nail (the circular part) is aligned with the 0-inch mark on the ruler.\n",
            "\n",
            "Step 2: Identify the ending point of the nail.\n",
            "The sharp tip of the nail extends past the 3-inch mark but does not reach the 4-inch mark.\n",
            "\n",
            "Step 3: Determine the length to the nearest inch.\n",
            "Since the tip is between 3 and 4 inches, and we are rounding to the nearest inch, I need to see which whole number it is closer to. The tip is clearly past 3 and before 4, so I need to estimate if it's closer to 3 or 4.\n",
            "\n",
            "Looking at the visual: the tip appears to be just shy of the 4-inch mark. It’s about 3.5 inches long (since it’s roughly halfway between 3 and 4), but since we need to round to the nearest inch, 3.5 rounds up to 4. However, let’s be precise. The nail's tip looks to be at approximately 3.2 to 3.4 inches. Since this is less than 3.5, it rounds down to 3 inches.\n",
            "\n",
            "But wait — let’s check again. The question says \"to the nearest inch\". The nail’s tip is clearly past 3 and is very close to 4, but not quite there. If it were exactly 3.5 inches, it would round to 4. But visually, it’s not 3.5 — it’s less than that. It appears to be about 3.3 inches. 3.3 is closer to 3 than to 4, so it rounds to 3.\n",
            "\n",
            "However, looking at the image more carefully, the tip appears to be just barely past the 3-inch mark and seems to be very close to the 4-inch mark — perhaps even a bit over 3.5. But since we are rounding to the nearest inch, we look at the midpoint between 3 and 4, which is 3.5. If the nail’s length is 3.5 inches or more, we round up to 4. If it’s less than 3.5, we round down to 3.\n",
            "\n",
            "In the image, the nail’s tip appears to be past 3.5. Let me estimate: from 3 to 4, the ruler is marked in 0.5 inch increments (since each major tick is 1 inch, and the midpoints are marked implicitly). The tip appears to be just before the 4-inch mark. So if it’s at 3.8 inches, that’s closer to 4 than to 3, so it rounds to 4. But if it’s at 3.2, it rounds to 3.\n",
            "\n",
            "Looking at the image, the tip is past the 3-inch mark and appears to be just shy of the 4-inch mark — maybe 3.7 or 3.8 inches? That would round to 4. However, the visual shows it's not quite to 4. Let’s consider: if the nail were exactly 3.5 inches, it would be at the midpoint, and typically, 3.5 rounds up to 4. But since it’s not exactly 3.5, let’s estimate.\n",
            "\n",
            "Actually, the nail’s tip is visually aligned very close to the 4-inch mark. It doesn't reach 4, but it's very close. For example, if it were 3.9 inches, it would round to 4. Since the question asks for the nearest inch, and the nail is clearly longer than 3 inches and very close to 4 inches, the answer should be 4 inches.\n",
            "\n",
            "But let’s be precise: I see the tip is clearly past the 3-inch mark and well before the 4-inch mark. It’s approximately 3.2 to 3.3 inches long. 3.2 is closer to 3 than to 4, so it should round to 3. 3.3 is still less than 3.5, so it rounds to 3.\n",
            "\n",
            "Wait, let me re-express: 3.2 is 0.2 away from 3, and 0.8 away from 4. So it’s closer to 3. 3.5 is exactly halfway, and conventionally rounds up. So 3.3 rounds to 3.\n",
            "\n",
            "But in the image, the tip appears to be very close to 3.5. It might be 3.4. 3.4 is still less than 3.5, so rounds to 3.\n",
            "\n",
            "But let’s look again. The ruler is marked at 0, 1, 2, 3,\n"
          ]
        }
      ],
      "source": [
        "image = train_dataset[100][\"image\"]\n",
        "prompt = train_dataset[100][\"prompt\"]\n",
        "\n",
        "inputs = tokenizer(\n",
        "    image,\n",
        "    prompt,\n",
        "    add_special_tokens = False,\n",
        "    return_tensors = \"pt\",\n",
        ").to(\"cuda\")\n",
        "\n",
        "from transformers import TextStreamer\n",
        "text_streamer = TextStreamer(tokenizer, skip_prompt = True)\n",
        "_ = model.generate(**inputs, streamer = text_streamer, max_new_tokens = 1024,\n",
        "                   use_cache = True, temperature = 1.0, min_p = 0.1)"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "Ux6iqP7z5YOo"
      },
      "source": [
        "<a name=\"Train\"></a>\n",
        "### Train the model\n",
        "\n",
        "Now set up the `GRPO` Trainer and all configurations! Note we actually enable `GSPO` as well!"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 18,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "ptqkXK2D4d6p",
        "outputId": "4a820765-b721-42d4-9e0c-f2675f74f1a1"
      },
      "outputs": [
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "Unsloth: We now expect `per_device_train_batch_size` to be a multiple of `num_generations`.\n",
            "We will change the batch size of 1 to the `num_generations` of 2\n"
          ]
        }
      ],
      "source": [
        "from trl import GRPOConfig, GRPOTrainer\n",
        "training_args = GRPOConfig(\n",
        "    learning_rate = 5e-6,\n",
        "    adam_beta1 = 0.9,\n",
        "    adam_beta2 = 0.99,\n",
        "    weight_decay = 0.1,\n",
        "    warmup_ratio = 0.1,\n",
        "    lr_scheduler_type = \"cosine\",\n",
        "    optim = \"adamw_8bit\",\n",
        "    logging_steps = 1,\n",
        "    log_completions = False,\n",
        "    per_device_train_batch_size = 1,\n",
        "    gradient_accumulation_steps = 1, # Increase to 4 for smoother training\n",
        "    num_generations = 2, # Decrease if out of memory\n",
        "    max_prompt_length = 1024,\n",
        "    max_completion_length = 1024,\n",
        "    num_train_epochs = 0.5, # Set to 1 for a full training run\n",
        "    # max_steps = 60,\n",
        "    save_steps = 60,\n",
        "    max_grad_norm = 0.1,\n",
        "    report_to = \"none\", # Can use Weights & Biases\n",
        "    output_dir = \"outputs\",\n",
        "\n",
        "    # Below enables GSPO:\n",
        "    importance_sampling_level = \"sequence\",\n",
        "    mask_truncated_completions = False,\n",
        "    loss_type = \"dr_grpo\",\n",
        ")"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "r9Mv8UZO5hz-"
      },
      "source": [
        "And let's run the trainer! If you scroll up, you'll see a table of rewards. The goal is to see the `reward` column increase!\n",
        "\n",
        "You might have to wait 150 to 200 steps for any action. You'll probably get 0 reward for the first 100 steps. Please be patient!\n",
        "\n",
        "| Step | Training Loss | reward    | reward_std | completion_length | kl       |\n",
        "|------|---------------|-----------|------------|-------------------|----------|\n",
        "| 1    | 0.000000      | 0.125000  | 0.000000   | 200.000000        | 0.000000 |\n",
        "| 2    | 0.000000      | 0.072375  | 0.248112   | 200.000000        | 0.000000 |\n",
        "| 3    | 0.000000      | -0.079000 | 0.163776   | 182.500000        | 0.000005 |\n",
        "\n",
        "During inference, you might encounter `addCriterion` or some weird gibberish outputs. Please read our [blog post](https://docs.unsloth.ai/new/vision-reinforcement-learning-vlm-rl#qwen-2.5-vl-vision-rl-issues-and-quirks) on why this occurs. It seems to be an inherent thing inside of the model, and we can ignore this."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 19,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/",
          "height": 1000
        },
        "id": "6fUaoYJEKgpb",
        "outputId": "e9aa22f1-aca6-4c80-d125-74ede220483d"
      },
      "outputs": [
        {
          "name": "stderr",
          "output_type": "stream",
          "text": [
            "The model is already on multiple devices. Skipping the move to device specified in `args`.\n",
            "==((====))==  Unsloth - 2x faster free finetuning | Num GPUs used = 1\n",
            "   \\\\   /|    Num examples = 566 | Num Epochs = 1 | Total steps = 60\n",
            "O^O/ \\_/ \\    Batch size per device = 2 | Gradient accumulation steps = 1\n",
            "\\        /    Data Parallel GPUs = 1 | Total batch size (2 x 1 x 1) = 2\n",
            " \"-____-\"     Trainable parameters = 43,646,976 of 8,810,770,672 (0.50% trained)\n"
          ]
        },
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the highest value on the X axis?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "30 \n",
            "Response:<REASONING>\n",
            "To determine the highest value on the X-axis of the given graph, I need to examine the horizontal axis (X-axis) which is labeled \"MICROGRAMS/ml-E-DNP-LYSINE-HCL\". The X-axis has tick marks with labeled values: 0, 5, 10, 15, 20, 25, and 30. The last tick mark on the right side of the axis is labeled \"30\". Therefore, the highest value shown on the X-axis is 30.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>30.0</SOLUTION>\n"
          ]
        },
        {
          "data": {
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              "\n",
              "    <div>\n",
              "      \n",
              "      <progress value='60' max='60' style='width:300px; height:20px; vertical-align: middle;'></progress>\n",
              "      [60/60 1:46:01, Epoch 0/1]\n",
              "    </div>\n",
              "    <table border=\"1\" class=\"dataframe\">\n",
              "  <thead>\n",
              " <tr style=\"text-align: left;\">\n",
              "      <th>Step</th>\n",
              "      <th>Training Loss</th>\n",
              "      <th>reward</th>\n",
              "      <th>reward_std</th>\n",
              "      <th>completions / mean_length</th>\n",
              "      <th>completions / min_length</th>\n",
              "      <th>completions / max_length</th>\n",
              "      <th>completions / clipped_ratio</th>\n",
              "      <th>completions / mean_terminated_length</th>\n",
              "      <th>completions / min_terminated_length</th>\n",
              "      <th>completions / max_terminated_length</th>\n",
              "      <th>kl</th>\n",
              "      <th>rewards / formatting_reward_func / mean</th>\n",
              "      <th>rewards / formatting_reward_func / std</th>\n",
              "      <th>rewards / correctness_reward_func / mean</th>\n",
              "      <th>rewards / correctness_reward_func / std</th>\n",
              "    </tr>\n",
              "  </thead>\n",
              "  <tbody>\n",
              "    <tr>\n",
              "      <td>1</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>155.500000</td>\n",
              "      <td>134.000000</td>\n",
              "      <td>177.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>155.500000</td>\n",
              "      <td>134.000000</td>\n",
              "      <td>177.000000</td>\n",
              "      <td>0.000016</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>2</td>\n",
              "      <td>-0.008600</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>0.707107</td>\n",
              "      <td>242.500000</td>\n",
              "      <td>230.000000</td>\n",
              "      <td>255.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>242.500000</td>\n",
              "      <td>230.000000</td>\n",
              "      <td>255.000000</td>\n",
              "      <td>0.000018</td>\n",
              "      <td>1.500000</td>\n",
              "      <td>0.707107</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
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              "    <tr>\n",
              "      <td>3</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>607.000000</td>\n",
              "      <td>555.000000</td>\n",
              "      <td>659.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>607.000000</td>\n",
              "      <td>555.000000</td>\n",
              "      <td>659.000000</td>\n",
              "      <td>0.000031</td>\n",
              "      <td>2.000000</td>\n",
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              "      <td>0.000000</td>\n",
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              "    <tr>\n",
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              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>388.000000</td>\n",
              "      <td>308.000000</td>\n",
              "      <td>468.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>388.000000</td>\n",
              "      <td>308.000000</td>\n",
              "      <td>468.000000</td>\n",
              "      <td>0.000009</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
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              "    <tr>\n",
              "      <td>5</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>347.000000</td>\n",
              "      <td>319.000000</td>\n",
              "      <td>375.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>347.000000</td>\n",
              "      <td>319.000000</td>\n",
              "      <td>375.000000</td>\n",
              "      <td>0.000020</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
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              "    <tr>\n",
              "      <td>6</td>\n",
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              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>303.000000</td>\n",
              "      <td>300.000000</td>\n",
              "      <td>306.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>303.000000</td>\n",
              "      <td>300.000000</td>\n",
              "      <td>306.000000</td>\n",
              "      <td>0.000008</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
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              "    <tr>\n",
              "      <td>7</td>\n",
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              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>317.500000</td>\n",
              "      <td>202.000000</td>\n",
              "      <td>433.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>317.500000</td>\n",
              "      <td>202.000000</td>\n",
              "      <td>433.000000</td>\n",
              "      <td>0.000008</td>\n",
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              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
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              "      <td>531.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>480.500000</td>\n",
              "      <td>430.000000</td>\n",
              "      <td>531.000000</td>\n",
              "      <td>0.000006</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
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              "    <tr>\n",
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              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>466.000000</td>\n",
              "      <td>367.000000</td>\n",
              "      <td>565.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>466.000000</td>\n",
              "      <td>367.000000</td>\n",
              "      <td>565.000000</td>\n",
              "      <td>0.000006</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
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              "    <tr>\n",
              "      <td>10</td>\n",
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              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>347.500000</td>\n",
              "      <td>323.000000</td>\n",
              "      <td>372.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>347.500000</td>\n",
              "      <td>323.000000</td>\n",
              "      <td>372.000000</td>\n",
              "      <td>0.000008</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>11</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>186.500000</td>\n",
              "      <td>132.000000</td>\n",
              "      <td>241.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>186.500000</td>\n",
              "      <td>132.000000</td>\n",
              "      <td>241.000000</td>\n",
              "      <td>0.000030</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>12</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>301.500000</td>\n",
              "      <td>282.000000</td>\n",
              "      <td>321.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>301.500000</td>\n",
              "      <td>282.000000</td>\n",
              "      <td>321.000000</td>\n",
              "      <td>0.000017</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>13</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>493.000000</td>\n",
              "      <td>426.000000</td>\n",
              "      <td>560.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>493.000000</td>\n",
              "      <td>426.000000</td>\n",
              "      <td>560.000000</td>\n",
              "      <td>0.000016</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>14</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>118.000000</td>\n",
              "      <td>116.000000</td>\n",
              "      <td>120.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>118.000000</td>\n",
              "      <td>116.000000</td>\n",
              "      <td>120.000000</td>\n",
              "      <td>0.000034</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>15</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>123.000000</td>\n",
              "      <td>113.000000</td>\n",
              "      <td>133.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>123.000000</td>\n",
              "      <td>113.000000</td>\n",
              "      <td>133.000000</td>\n",
              "      <td>0.000006</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>16</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>252.500000</td>\n",
              "      <td>228.000000</td>\n",
              "      <td>277.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>252.500000</td>\n",
              "      <td>228.000000</td>\n",
              "      <td>277.000000</td>\n",
              "      <td>0.000007</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>17</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>379.000000</td>\n",
              "      <td>360.000000</td>\n",
              "      <td>398.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>379.000000</td>\n",
              "      <td>360.000000</td>\n",
              "      <td>398.000000</td>\n",
              "      <td>0.000004</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>18</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>261.500000</td>\n",
              "      <td>209.000000</td>\n",
              "      <td>314.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>261.500000</td>\n",
              "      <td>209.000000</td>\n",
              "      <td>314.000000</td>\n",
              "      <td>0.000022</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>19</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000007</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>20</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000010</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>21</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>292.500000</td>\n",
              "      <td>205.000000</td>\n",
              "      <td>380.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>292.500000</td>\n",
              "      <td>205.000000</td>\n",
              "      <td>380.000000</td>\n",
              "      <td>0.000017</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>22</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>296.000000</td>\n",
              "      <td>225.000000</td>\n",
              "      <td>367.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>296.000000</td>\n",
              "      <td>225.000000</td>\n",
              "      <td>367.000000</td>\n",
              "      <td>0.000042</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>23</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>282.500000</td>\n",
              "      <td>213.000000</td>\n",
              "      <td>352.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>282.500000</td>\n",
              "      <td>213.000000</td>\n",
              "      <td>352.000000</td>\n",
              "      <td>0.000023</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>24</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>188.500000</td>\n",
              "      <td>182.000000</td>\n",
              "      <td>195.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>188.500000</td>\n",
              "      <td>182.000000</td>\n",
              "      <td>195.000000</td>\n",
              "      <td>0.000006</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>25</td>\n",
              "      <td>0.222000</td>\n",
              "      <td>1.000000</td>\n",
              "      <td>1.414214</td>\n",
              "      <td>702.500000</td>\n",
              "      <td>381.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>0.500000</td>\n",
              "      <td>381.000000</td>\n",
              "      <td>381.000000</td>\n",
              "      <td>381.000000</td>\n",
              "      <td>0.000007</td>\n",
              "      <td>1.000000</td>\n",
              "      <td>1.414214</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>26</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>265.000000</td>\n",
              "      <td>245.000000</td>\n",
              "      <td>285.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>265.000000</td>\n",
              "      <td>245.000000</td>\n",
              "      <td>285.000000</td>\n",
              "      <td>0.000006</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>27</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>142.500000</td>\n",
              "      <td>119.000000</td>\n",
              "      <td>166.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>142.500000</td>\n",
              "      <td>119.000000</td>\n",
              "      <td>166.000000</td>\n",
              "      <td>0.000034</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>28</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>237.000000</td>\n",
              "      <td>218.000000</td>\n",
              "      <td>256.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>237.000000</td>\n",
              "      <td>218.000000</td>\n",
              "      <td>256.000000</td>\n",
              "      <td>0.000013</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>29</td>\n",
              "      <td>0.176400</td>\n",
              "      <td>1.000000</td>\n",
              "      <td>1.414214</td>\n",
              "      <td>768.500000</td>\n",
              "      <td>513.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>0.500000</td>\n",
              "      <td>513.000000</td>\n",
              "      <td>513.000000</td>\n",
              "      <td>513.000000</td>\n",
              "      <td>0.000008</td>\n",
              "      <td>1.000000</td>\n",
              "      <td>1.414214</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>30</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>131.000000</td>\n",
              "      <td>131.000000</td>\n",
              "      <td>131.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>131.000000</td>\n",
              "      <td>131.000000</td>\n",
              "      <td>131.000000</td>\n",
              "      <td>0.000011</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>31</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>424.500000</td>\n",
              "      <td>405.000000</td>\n",
              "      <td>444.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>424.500000</td>\n",
              "      <td>405.000000</td>\n",
              "      <td>444.000000</td>\n",
              "      <td>0.000064</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>32</td>\n",
              "      <td>0.187100</td>\n",
              "      <td>1.000000</td>\n",
              "      <td>1.414214</td>\n",
              "      <td>753.000000</td>\n",
              "      <td>482.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>0.500000</td>\n",
              "      <td>482.000000</td>\n",
              "      <td>482.000000</td>\n",
              "      <td>482.000000</td>\n",
              "      <td>0.000010</td>\n",
              "      <td>1.000000</td>\n",
              "      <td>1.414214</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>33</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>411.500000</td>\n",
              "      <td>359.000000</td>\n",
              "      <td>464.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>411.500000</td>\n",
              "      <td>359.000000</td>\n",
              "      <td>464.000000</td>\n",
              "      <td>0.000013</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>34</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>176.500000</td>\n",
              "      <td>161.000000</td>\n",
              "      <td>192.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>176.500000</td>\n",
              "      <td>161.000000</td>\n",
              "      <td>192.000000</td>\n",
              "      <td>0.000009</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>35</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>253.000000</td>\n",
              "      <td>228.000000</td>\n",
              "      <td>278.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>253.000000</td>\n",
              "      <td>228.000000</td>\n",
              "      <td>278.000000</td>\n",
              "      <td>0.000034</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>36</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>408.000000</td>\n",
              "      <td>357.000000</td>\n",
              "      <td>459.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>408.000000</td>\n",
              "      <td>357.000000</td>\n",
              "      <td>459.000000</td>\n",
              "      <td>0.000006</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>37</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>199.500000</td>\n",
              "      <td>194.000000</td>\n",
              "      <td>205.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>199.500000</td>\n",
              "      <td>194.000000</td>\n",
              "      <td>205.000000</td>\n",
              "      <td>0.000016</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>38</td>\n",
              "      <td>0.068000</td>\n",
              "      <td>1.000000</td>\n",
              "      <td>1.414214</td>\n",
              "      <td>925.500000</td>\n",
              "      <td>827.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>0.500000</td>\n",
              "      <td>827.000000</td>\n",
              "      <td>827.000000</td>\n",
              "      <td>827.000000</td>\n",
              "      <td>0.000012</td>\n",
              "      <td>1.000000</td>\n",
              "      <td>1.414214</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>39</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000004</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>40</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>105.500000</td>\n",
              "      <td>105.000000</td>\n",
              "      <td>106.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>105.500000</td>\n",
              "      <td>105.000000</td>\n",
              "      <td>106.000000</td>\n",
              "      <td>0.000014</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>41</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>223.000000</td>\n",
              "      <td>190.000000</td>\n",
              "      <td>256.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>223.000000</td>\n",
              "      <td>190.000000</td>\n",
              "      <td>256.000000</td>\n",
              "      <td>0.000011</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>42</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>225.500000</td>\n",
              "      <td>193.000000</td>\n",
              "      <td>258.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>225.500000</td>\n",
              "      <td>193.000000</td>\n",
              "      <td>258.000000</td>\n",
              "      <td>0.000057</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>43</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>256.500000</td>\n",
              "      <td>201.000000</td>\n",
              "      <td>312.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>256.500000</td>\n",
              "      <td>201.000000</td>\n",
              "      <td>312.000000</td>\n",
              "      <td>0.000008</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>44</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000017</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>45</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>301.500000</td>\n",
              "      <td>251.000000</td>\n",
              "      <td>352.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>301.500000</td>\n",
              "      <td>251.000000</td>\n",
              "      <td>352.000000</td>\n",
              "      <td>0.000010</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>46</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>199.500000</td>\n",
              "      <td>187.000000</td>\n",
              "      <td>212.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>199.500000</td>\n",
              "      <td>187.000000</td>\n",
              "      <td>212.000000</td>\n",
              "      <td>0.000011</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>47</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>184.000000</td>\n",
              "      <td>183.000000</td>\n",
              "      <td>185.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>184.000000</td>\n",
              "      <td>183.000000</td>\n",
              "      <td>185.000000</td>\n",
              "      <td>0.000005</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>48</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>341.000000</td>\n",
              "      <td>336.000000</td>\n",
              "      <td>346.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>341.000000</td>\n",
              "      <td>336.000000</td>\n",
              "      <td>346.000000</td>\n",
              "      <td>0.000009</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>49</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>342.500000</td>\n",
              "      <td>299.000000</td>\n",
              "      <td>386.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>342.500000</td>\n",
              "      <td>299.000000</td>\n",
              "      <td>386.000000</td>\n",
              "      <td>0.000012</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>50</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>387.000000</td>\n",
              "      <td>352.000000</td>\n",
              "      <td>422.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>387.000000</td>\n",
              "      <td>352.000000</td>\n",
              "      <td>422.000000</td>\n",
              "      <td>0.000018</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>51</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>154.500000</td>\n",
              "      <td>149.000000</td>\n",
              "      <td>160.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>154.500000</td>\n",
              "      <td>149.000000</td>\n",
              "      <td>160.000000</td>\n",
              "      <td>0.000007</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "    </tr>\n",
              "    <tr>\n",
              "      <td>52</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>337.500000</td>\n",
              "      <td>304.000000</td>\n",
              "      <td>371.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>337.500000</td>\n",
              "      <td>304.000000</td>\n",
              "      <td>371.000000</td>\n",
              "      <td>0.000002</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
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              "    <tr>\n",
              "      <td>53</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>4.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>381.000000</td>\n",
              "      <td>337.000000</td>\n",
              "      <td>425.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>381.000000</td>\n",
              "      <td>337.000000</td>\n",
              "      <td>425.000000</td>\n",
              "      <td>0.000010</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
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              "    <tr>\n",
              "      <td>54</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1024.000000</td>\n",
              "      <td>1.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000008</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
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              "    <tr>\n",
              "      <td>55</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>300.000000</td>\n",
              "      <td>211.000000</td>\n",
              "      <td>389.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>300.000000</td>\n",
              "      <td>211.000000</td>\n",
              "      <td>389.000000</td>\n",
              "      <td>0.000012</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>0.000000</td>\n",
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              "    <tr>\n",
              "      <td>56</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>2.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>235.000000</td>\n",
              "      <td>191.000000</td>\n",
              "      <td>279.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>235.000000</td>\n",
              "      <td>191.000000</td>\n",
              "      <td>279.000000</td>\n",
              "      <td>0.000012</td>\n",
              "      <td>2.000000</td>\n",
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              "      <td>188.500000</td>\n",
              "      <td>142.000000</td>\n",
              "      <td>235.000000</td>\n",
              "      <td>0.000000</td>\n",
              "      <td>188.500000</td>\n",
              "      <td>142.000000</td>\n",
              "      <td>235.000000</td>\n",
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              "      <td>2.000000</td>\n",
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              "      <td>0.000000</td>\n",
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              "    <tr>\n",
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              "      <td>285.000000</td>\n",
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              "      <td>1024.000000</td>\n",
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            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the age gap between these two people in image?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "6 \n",
            "Response:<REASONING>\n",
            "To determine the age gap between the two people in the image, I would need to analyze visual cues such as facial features, body proportions, hairstyle, clothing style, and context (e.g., era of the photo). However, this is a black-and-white photograph, and without clear identifiers like birth dates, names, or other contextual information (e.g., from a caption or historical record), it is impossible to accurately determine their actual ages or the age difference between them.\n",
            "\n",
            "The image shows two men standing next to a makeshift hut. They appear to be young adults, possibly in their late teens to early twenties, judging by their facial features and attire. However, without knowing their exact birth years or any other reference point, I cannot calculate a numerical age gap.\n",
            "\n",
            "Therefore, based on the information available in the image alone, it is not possible to solve the question.\n",
            "\n",
            "<REASONING>\n",
            "The question cannot be solved with the given information. There is no data to determine the ages of the two individuals or any reliable way to infer their age difference from the image.\n",
            "\n",
            "<SOLUTION>\n",
            "0.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Subtract all yellow metallic balls. Subtract all small yellow shiny things. How many objects are left?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "8 \n",
            "Response:<REASONING>\n",
            "We are given a set of 10 objects in the image. Let's identify and categorize them first:\n",
            "\n",
            "1.  A large blue cube (center)\n",
            "2.  A small purple cylinder (left of center)\n",
            "3.  A small yellow cylinder (behind the purple cylinder)\n",
            "4.  A small gold/yellow metallic ball (far left)\n",
            "5.  A small purple cylinder (behind the large blue cube)\n",
            "6.  A small blue cube (behind the large blue cube)\n",
            "7.  A small cyan cube (right side)\n",
            "8.  A small blue cube (right side, in front of the cyan cube)\n",
            "9.  A small gray sphere (behind the large blue cube)\n",
            "10. A small silver/grey metallic sphere (in front of the large blue cube)\n",
            "\n",
            "Now, we need to subtract objects based on the conditions:\n",
            "\n",
            "- \"Subtract all yellow metallic balls\": There is one yellow metallic ball (the gold/yellow one on the far left). We subtract this.\n",
            "\n",
            "- \"Subtract all small yellow shiny things\": This includes the small yellow cylinder (which is shiny) and the small yellow metallic ball (which we've already subtracted). So we subtract the small yellow cylinder.\n",
            "\n",
            "After subtracting these 2 objects, we are left with 8 objects.\n",
            "\n",
            "Let's list the remaining objects:\n",
            "- The large blue cube\n",
            "- The small purple cylinder (left of center)\n",
            "- The small blue cube (behind the large blue cube)\n",
            "- The small cyan cube\n",
            "- The small blue cube (right side)\n",
            "- The small gray sphere\n",
            "- The small silver/grey metallic sphere\n",
            "- The small purple cylinder (behind the large blue cube) — wait, this is the same as the one we already listed? Let me check.\n",
            "\n",
            "Actually, let's re-express the original list with indices for clarity:\n",
            "\n",
            "Original 10 objects:\n",
            "1. Large blue cube\n",
            "2. Small purple cylinder (left)\n",
            "3. Small yellow cylinder\n",
            "4. Small gold/yellow metallic ball\n",
            "5. Small purple cylinder (behind large blue cube) — this is a second purple cylinder\n",
            "6. Small blue cube\n",
            "7. Small cyan cube\n",
            "8. Small blue cube (right)\n",
            "9. Small gray sphere\n",
            "10. Small silver/grey metallic sphere\n",
            "\n",
            "So, we subtract:\n",
            "- Object 4 (yellow metallic ball)\n",
            "- Object 3 (small yellow cylinder)\n",
            "\n",
            "That leaves us with 8 objects: 1, 2, 5, 6, 7, 8, 9, 10.\n",
            "\n",
            "Therefore, 8 objects are left.\n",
            "</REASONING>\n",
            "\n",
            "<SOLUTION>8.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Subtract all tiny balls. Subtract all green metallic things. How many objects are left?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "5 \n",
            "Response:<REASONING>\n",
            "I need to identify all objects in the image and then apply the two subtraction rules.\n",
            "\n",
            "Step 1: List all objects in the image.\n",
            "There are 10 objects total:\n",
            "1. Large green metallic cylinder\n",
            "2. Small yellow cylinder\n",
            "3. Large yellow cube\n",
            "4. Small cyan cube\n",
            "5. Small silver cube\n",
            "6. Small gold metallic sphere (tiny ball)\n",
            "7. Small brown sphere (tiny ball)\n",
            "8. Small green sphere (tiny ball)\n",
            "9. Large red sphere\n",
            "10. Small purple sphere (tiny ball)\n",
            "\n",
            "Step 2: Apply the first instruction: \"Subtract all tiny balls\".\n",
            "Tiny balls are: the small gold sphere, the small brown sphere, the small green sphere, and the small purple sphere.\n",
            "That's 4 objects to subtract.\n",
            "\n",
            "Remaining objects after removing tiny balls: \n",
            "- Large green metallic cylinder\n",
            "- Small yellow cylinder\n",
            "- Large yellow cube\n",
            "- Small cyan cube\n",
            "- Small silver cube\n",
            "- Large red sphere\n",
            "\n",
            "That's 6 objects left.\n",
            "\n",
            "Step 3: Apply the second instruction: \"Subtract all green metallic things\".\n",
            "The large green metallic cylinder is the only green metallic thing.\n",
            "Subtracting that leaves:\n",
            "- Small yellow cylinder\n",
            "- Large yellow cube\n",
            "- Small cyan cube\n",
            "- Small silver cube\n",
            "- Large red sphere\n",
            "\n",
            "That's 5 objects left.\n",
            "\n",
            "Therefore, after applying both subtractions, 5 objects remain.\n",
            "</REASONING>\n",
            "<SOLUTION>5.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the highest accuracy reported in the whole chart?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "8 \n",
            "Response:<REASONING>\n",
            "To find the highest accuracy reported in the whole chart, I need to:\n",
            "\n",
            "1. Examine each bar in the grouped horizontal bar chart.\n",
            "2. Read the value on the x-axis (labeled \"Accuracy\") corresponding to the end of each bar.\n",
            "3. Identify the bar that extends the furthest to the right, as that represents the highest accuracy value.\n",
            "4. The categories on the y-axis are \"dash\", \"port\", \"muzzle\", \"want\", and \"breed\", and each has four bars representing \"bow\", \"swing\", \"proud\", and \"soup\".\n",
            "\n",
            "Looking at the chart:\n",
            "- For \"dash\": the \"swing\" bar (yellow-orange) goes up to about 6.5.\n",
            "- For \"port\": the \"bow\" bar (red) goes up to about 7.5.\n",
            "- For \"muzzle\": the \"bow\" bar (red) goes up to about 4.5.\n",
            "- For \"want\": the \"swing\" bar (yellow-orange) goes up to about 7.5.\n",
            "- For \"breed\": the \"bow\" bar (red) goes up to about 6.5.\n",
            "\n",
            "The highest bar appears to be for \"port\" (bow) and \"want\" (swing), both reaching approximately 7.5. Since the question asks for the highest accuracy reported, and these are tied at 7.5, I will take 7.5 as the answer.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>7.5</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Subtract all large red rubber blocks. Subtract all tiny red matte objects. How many objects are left?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "2 \n",
            "Response:<REASONING>\n",
            "Let's analyze the objects in the image step by step.\n",
            "\n",
            "1. Identify all objects:\n",
            "   - There is a large, shiny red cube (metallic, not rubber).\n",
            "   - There is a small yellow cube (matte, not red).\n",
            "   - There is a small red cube (matte, appears to be the \"tiny red matte object\").\n",
            "\n",
            "2. Apply the subtraction rules:\n",
            "\n",
            "   a. \"Subtract all large red rubber blocks\":\n",
            "      - The large red cube is red and large, but it is shiny/metallic, not rubber. So, it is NOT a \"rubber block\". Therefore, it is NOT subtracted.\n",
            "      - The small red cube is not large, so it is not subtracted.\n",
            "      - The yellow cube is not red, so it is not subtracted.\n",
            "      - So, no objects are subtracted in this step.\n",
            "\n",
            "   b. \"Subtract all tiny red matte objects\":\n",
            "      - The small red cube is tiny, red, and matte. This matches exactly.\n",
            "      - So, we subtract this small red cube.\n",
            "\n",
            "3. Count remaining objects:\n",
            "   - Large shiny red cube: remains (not subtracted in step a)\n",
            "   - Small yellow cube: remains (not subtracted in either step)\n",
            "   - Small red matte cube: subtracted (step b)\n",
            "\n",
            "   So, 2 objects remain.\n",
            "\n",
            "Therefore, the answer is 2.\n",
            "</REASONING>\n",
            "<SOLUTION>2.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the value of the largest bar?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "10000 \n",
            "Response:<REASONING>\n",
            "To determine the value of the largest bar in the given bar chart, I will follow these steps:\n",
            "\n",
            "1. Examine the chart to identify the categories on the x-axis: \"picnic\", \"fair\", \"draw\", and \"candle\".\n",
            "2. Observe the height of each bar and compare it against the y-axis, which is on a logarithmic scale (powers of 10: 10^2, 10^4, 10^6, 10^8, 10^10).\n",
            "3. The tallest bar corresponds to the category \"fair\".\n",
            "4. The top of the \"fair\" bar aligns exactly with the 10^4 mark on the y-axis.\n",
            "5. Therefore, the value of the largest bar is 10^4, which equals 10,000.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>10000.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>如图，AB、BC、CD、DA都是⊙O的切线，已知AD＝2，BC＝5，则AB+CD的值是（）. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "7 \n",
            "Response:<REASONING>\n",
            "We are given a quadrilateral ABCD that is circumscribed about a circle with center O. This means that all four sides (AB, BC, CD, DA) are tangent to the circle.\n",
            "\n",
            "A key property of tangents drawn from a common external point to a circle is that they are equal in length.\n",
            "\n",
            "Let’s denote the points of tangency:\n",
            "- Let the point of tangency on AB be P.\n",
            "- Let the point of tangency on BC be Q.\n",
            "- Let the point of tangency on CD be R.\n",
            "- Let the point of tangency on DA be S.\n",
            "\n",
            "Then, by the tangent segment theorem:\n",
            "- AP = AS (tangents from point A)\n",
            "- BP = BQ (tangents from point B)\n",
            "- CQ = CR (tangents from point C)\n",
            "- DR = DS (tangents from point D)\n",
            "\n",
            "Now, let’s express the sides of the quadrilateral in terms of these tangent segments:\n",
            "- AB = AP + BP = AS + BQ\n",
            "- BC = BQ + CQ\n",
            "- CD = CR + DR = CQ + DS\n",
            "- DA = DS + AS\n",
            "\n",
            "We are given:\n",
            "- AD = DA = 2\n",
            "- BC = 5\n",
            "\n",
            "We need to find AB + CD.\n",
            "\n",
            "Let’s write AB + CD:\n",
            "AB + CD = (AS + BQ) + (CQ + DS)\n",
            "\n",
            "Group the terms:\n",
            "= (AS + DS) + (BQ + CQ)\n",
            "\n",
            "But AS + DS = AD = 2 (since AS and DS are the two tangent segments from point A and D to the circle, and they form side AD).\n",
            "\n",
            "Also, BQ + CQ = BC = 5 (since BQ and CQ are the two tangent segments from point B and C to the circle, and they form side BC).\n",
            "\n",
            "Therefore:\n",
            "AB + CD = 2 + 5 = 7\n",
            "\n",
            "Alternatively, we can see that:\n",
            "AB + CD = (AS + BQ) + (CQ + DS) = (AS + DS) + (BQ + CQ) = AD + BC = 2 + 5 = 7\n",
            "\n",
            "This is a direct application of the property that for any tangential quadrilateral (a quadrilateral with an incircle), the sums of the lengths of opposite sides are equal. That is, AB + CD = AD + BC.\n",
            "\n",
            "So, AB + CD = AD + BC = 2 + 5 = 7.\n",
            "\n",
            "</REASONING>\n",
            "\n",
            "<SOLUTION>7.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the age gap between these two people in image?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "16 \n",
            "Response:<REASONING>\n",
            "To determine the age gap between the two individuals in the image, I need to identify who they are and their respective birth years.\n",
            "\n",
            "1. The person on the right is Barack Obama, the 44th President of the United States. He was born on August 4, 1961.\n",
            "2. The person on the left is Aung San Suu Kyi, a Burmese politician and Nobel laureate. She was born on June 19, 1945.\n",
            "\n",
            "Next, I calculate the age difference by subtracting their birth years:\n",
            "\n",
            "Age gap = Birth year of older person - Birth year of younger person\n",
            "Since 1961 is after 1945, Barack Obama was born later, meaning Aung San Suu Kyi is older.\n",
            "\n",
            "Age gap = 1961 - 1945 = 16 years.\n",
            "\n",
            "However, since they were born in different months and years, I should check if the exact age difference is 16 years or if it's slightly less due to the dates.\n",
            "\n",
            "- Aung San Suu Kyi was born in June 1945.\n",
            "- Barack Obama was born in August 1961.\n",
            "\n",
            "So, when Obama was born, Suu Kyi was 16 years and 1 month old (since June 1945 to August 1961 is 16 years and 2 months). Therefore, the age gap is 16 years and 2 months.\n",
            "\n",
            "But since the question asks for a single float value, and typically in such contexts, we report the whole number of years, the age gap is 16 years.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>16.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>How many algorithms have accuracy lower than 8 in at least one dataset?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "3 \n",
            "Response:<REASONING>\n",
            "To solve this question, we need to examine the bar chart and determine for each algorithm (dataset) whether its accuracy is lower than 8 in at least one of the two datasets shown (‘crest’ and ‘louder’).\n",
            "\n",
            "The datasets shown are:\n",
            "- cork\n",
            "- lover\n",
            "- cook\n",
            "\n",
            "For each dataset, we look at the heights of the two bars (blue for ‘crest’, green for ‘louder’).\n",
            "\n",
            "1. For ‘cork’:\n",
            "   - ‘crest’ (blue bar) has accuracy = 3 (which is < 8)\n",
            "   - ‘louder’ (green bar) has accuracy = 8 (which is not < 8)\n",
            "   → Since at least one accuracy is < 8, ‘cork’ qualifies.\n",
            "\n",
            "2. For ‘lover’:\n",
            "   - ‘crest’ (blue bar) has accuracy = 6 (which is < 8)\n",
            "   - ‘louder’ (green bar) has accuracy = 5 (which is < 8)\n",
            "   → Since both are < 8, ‘lover’ qualifies.\n",
            "\n",
            "3. For ‘cook’:\n",
            "   - ‘crest’ (blue bar) has accuracy = 7 (which is < 8)\n",
            "   - ‘louder’ (green bar) has accuracy = 9 (which is not < 8)\n",
            "   → Since at least one accuracy is < 8, ‘cook’ qualifies.\n",
            "\n",
            "Therefore, all three algorithms (cork, lover, cook) have at least one dataset where accuracy is lower than 8.\n",
            "\n",
            "The question asks for “how many algorithms”, so we count the number of distinct algorithms (i.e., distinct x-axis labels), which is 3.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>3.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the age gap between these two people in image?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "12 \n",
            "Response:<REASONING>\n",
            "To determine the age gap between the two people in the image, I would need to know their exact ages. However, since this is a photograph of two professional football players (one from Fulham FC wearing a white kit, and one from Chelsea FC wearing a blue kit), and no birth dates or ages are provided in the image or in any accompanying text, it is impossible to calculate an exact age difference.\n",
            "\n",
            "In real-world scenarios, one might attempt to estimate their ages by:\n",
            "1. Identifying the players from the image (e.g., using their appearance, known career timelines, or public records).\n",
            "2. Looking up their birth dates from official sources (e.g., club websites, sports databases like Transfermarkt or Wikipedia).\n",
            "3. Calculating the difference between their birth years.\n",
            "\n",
            "But since this is a hypothetical or image-only question without additional data, and no player identification or birth dates are given, the age gap cannot be determined. The question as posed lacks sufficient information to compute a numerical answer.\n",
            "\n",
            "Therefore, the most accurate response is that the age gap cannot be determined from the given information.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>0.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the age gap between these two people in image?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "3 \n",
            "Response:<REASONING>\n",
            "To estimate the age gap between the two individuals in the image, I would need to make educated guesses about their ages based on their physical appearance, such as facial features, hair, and skin texture, which are common indicators of age. Since I cannot determine their exact birth years from the image, I will estimate their approximate ages.\n",
            "\n",
            "Looking at the man on the left (in the suit):\n",
            "- He has a graying beard and hair, and visible wrinkles, suggesting he is in his late 50s to early 70s. I will estimate his age to be around 65.\n",
            "\n",
            "Looking at the man on the right (in the military uniform):\n",
            "- He appears to be in his late 50s to early 60s. His hair is graying but still relatively full, and his skin shows some signs of aging but not as pronounced as the man on the left. I will estimate his age to be around 60.\n",
            "\n",
            "Age gap = Age of person on the left - Age of person on the right\n",
            "Age gap = 65 - 60 = 5 years.\n",
            "\n",
            "This is an estimation based on visual cues, and actual ages could vary. However, given the appearance, the most reasonable estimate for the age gap is around 5 years.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>5.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>For Group C, in which week is the cumulative increase in weight , the highest?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "3 \n",
            "Response:<REASONING>\n",
            "To determine the week in which the cumulative increase in weight is the highest for Group C, I will:\n",
            "\n",
            "1. Identify the line corresponding to Group C on the graph. The label \"GROUP C CONTROL\" is located near the lower right portion of the graph, and its line is dotted.\n",
            "\n",
            "2. Examine the y-axis (Cumulative Increase in Weight in Grams) and the x-axis (Weeks) to understand the scale and units.\n",
            "\n",
            "3. Trace along the dotted line for Group C across the weeks (0 through 6) and note the cumulative weight increase at each week.\n",
            "\n",
            "4. Compare the values at each week to find the maximum.\n",
            "\n",
            "From the graph:\n",
            "- At Week 0: The weight increase is 0 grams.\n",
            "- At Week 1: The weight increase is approximately 50 grams.\n",
            "- At Week 2: The weight increase is approximately 100 grams.\n",
            "- At Week 3: The weight increase is approximately 200 grams.\n",
            "- At Week 4: The weight increase is approximately 100 grams.\n",
            "- At Week 5: The weight increase is approximately 150 grams.\n",
            "- At Week 6: The weight increase is approximately 200 grams.\n",
            "\n",
            "Comparing these values, the highest cumulative increase for Group C occurs at Week 3 and Week 6, both at approximately 200 grams. Since the question asks for \"the week\", and both weeks have the same highest value, I should choose the first occurrence or the one with the highest value if there is a tie. However, looking at the graph more carefully, at Week 3 the point is at 200 grams, and at Week 6 it is also at 200 grams. But since the question likely expects a single answer, and Week 3 is the first time it reaches 200 grams, I will select Week 3. However, if the graph shows Week 6 is slightly higher, I would select Week 6. Upon closer inspection of the graph, the point at Week 6 appears to be slightly higher than 200 grams. The point at Week 3 is exactly at 200 grams, and the point at Week 6 is visibly higher than 200 grams. Let me re-examine the graph.\n",
            "\n",
            "Actually, looking at the graph more precisely:\n",
            "- Week 3: The point is at 200 grams.\n",
            "- Week 6: The point is above 200 grams, appearing to be approximately 220 grams.\n",
            "\n",
            "Therefore, the highest cumulative increase for Group C is at Week 6.\n",
            "\n",
            "</REASONING>\n",
            "\n",
            "<SOLUTION>6.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>A food industry researcher compiled the revenues of several pizzerias. How much did Dan's Deep Dish make from pizza sales? (Unit: $). Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "22 \n",
            "Response:<REASONING>\n",
            "To determine how much Dan's Deep Dish made from pizza sales, I need to sum the revenues from its two types of pizzas: Pepperoni pizzas and Supreme pizzas.\n",
            "\n",
            "From the table:\n",
            "- Dan's Deep Dish sold Pepperoni pizzas for $8.\n",
            "- Dan's Deep Dish sold Supreme pizzas for $14.\n",
            "\n",
            "So, total revenue = $8 + $14 = $22.\n",
            "\n",
            "Therefore, Dan's Deep Dish made $22 from pizza sales.\n",
            "</REASONING>\n",
            "<SOLUTION>\n",
            "22.0\n",
            "</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>An author recorded how many words she wrote in the past 3 days. How many words in total did the author write on Thursday and Friday? (Unit: words). Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "679 \n",
            "Response:<REASONING>\n",
            "To find the total number of words the author wrote on Thursday and Friday, I need to:\n",
            "1. Identify the number of words written on Thursday from the table: 374 words.\n",
            "2. Identify the number of words written on Friday from the table: 305 words.\n",
            "3. Add these two numbers together to get the total.\n",
            "Calculation: 374 + 305 = 679.\n",
            "</REASONING>\n",
            "<SOLUTION>\n",
            "679.0\n",
            "</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the value of the smallest bar?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "1 \n",
            "Response:<REASONING>\n",
            "To determine the value of the smallest bar in the given bar chart, I will follow these steps:\n",
            "\n",
            "1. Examine the bar chart and identify the labels on the y-axis: \"twist\", \"priest\", \"fringe\", \"pan\", and \"curse\".\n",
            "2. Observe the length of each bar along the x-axis, which represents the \"Values\".\n",
            "3. The x-axis has tick marks at 0, 2, 4, 6, 8, and 10, indicating the scale.\n",
            "4. Estimate the value for each bar by seeing how far it extends from the origin (0) along the x-axis:\n",
            "   - \"twist\": The bar extends to approximately 1.\n",
            "   - \"priest\": The bar extends to approximately 9.\n",
            "   - \"fringe\": The bar extends to approximately 4.\n",
            "   - \"pan\": The bar extends to approximately 2.\n",
            "   - \"curse\": The bar extends to approximately 4.\n",
            "5. Compare these values: 1, 9, 4, 2, 4.\n",
            "6. The smallest value among these is 1, corresponding to the \"twist\" bar.\n",
            "\n",
            "Therefore, the value of the smallest bar is 1.\n",
            "</REASONING>\n",
            "<SOLUTION>1.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>How many items sold more than 3 units in at least one store?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "7 \n",
            "Response:<REASONING>\n",
            "To solve this question, I need to examine the bar chart and count how many items (categories on the x-axis) have at least one store (wrist or cult) where the number of units sold is greater than 3.\n",
            "\n",
            "Let me go through each item:\n",
            "\n",
            "1. **deputy**: \n",
            "   - wrist (red): 6 units (>3) ✓\n",
            "   - cult (blue): 3 units (=3, not >3)\n",
            "   → Since at least one store sold >3, this item counts.\n",
            "\n",
            "2. **brave**:\n",
            "   - wrist (red): 7 units (>3) ✓\n",
            "   - cult (blue): 4 units (>3) ✓\n",
            "   → This item counts.\n",
            "\n",
            "3. **violin**:\n",
            "   - wrist (red): 9 units (>3) ✓\n",
            "   - cult (blue): 9 units (>3) ✓\n",
            "   → This item counts.\n",
            "\n",
            "4. **arch**:\n",
            "   - wrist (red): 7 units (>3) ✓\n",
            "   - cult (blue): 5 units (>3) ✓\n",
            "   → This item counts.\n",
            "\n",
            "5. **breeze**:\n",
            "   - wrist (red): 4 units (>3) ✓\n",
            "   - cult (blue): 1 unit (<3)\n",
            "   → This item counts.\n",
            "\n",
            "6. **above**:\n",
            "   - wrist (red): 9 units (>3) ✓\n",
            "   - cult (blue): 9 units (>3) ✓\n",
            "   → This item counts.\n",
            "\n",
            "7. **boom**:\n",
            "   - wrist (red): 6 units (>3) ✓\n",
            "   - cult (blue): 3 units (=3, not >3)\n",
            "   → This item counts.\n",
            "\n",
            "So, all 7 items have at least one store where more than 3 units were sold.\n",
            "\n",
            "Therefore, the answer is 7.\n",
            "</REASONING>\n",
            "<SOLUTION>7.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the global maximum of this function?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "4 \n",
            "Response:<REASONING>\n",
            "To find the global maximum of the given function from its graph, I need to:\n",
            "\n",
            "1. Understand what \"global maximum\" means: It is the highest value that the function attains over its entire domain (in this case, from x = -3π/2 to x = 2π, and likely beyond, but we observe the peaks within this interval).\n",
            "\n",
            "2. Observe the graph:\n",
            "   - The function is periodic (repeats its pattern), with peaks and troughs.\n",
            "   - The graph shows several peaks (local maxima) and troughs (local minima).\n",
            "   - The highest points on the graph are the peaks.\n",
            "   - Looking at the y-axis, the peaks appear to reach a height of 4.\n",
            "   - There is a horizontal dashed line at y = 1, which appears to be a horizontal asymptote or a reference line, but it is not the maximum.\n",
            "\n",
            "3. Identify the highest point(s):\n",
            "   - The graph has peaks at approximately x = -π/2, π/2, 3π/2, etc.\n",
            "   - The y-value at these peaks is 4.\n",
            "   - Since the function reaches 4 and no point on the graph goes above 4, this is the global maximum.\n",
            "\n",
            "4. Confirm: The graph shows the function oscillating between approximately -2 and 4. The highest value reached is clearly 4.\n",
            "\n",
            "Therefore, the global maximum is 4.\n",
            "</REASONING>\n",
            "\n",
            "<SOLUTION>4.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>As shown in the figure, E is any point in ▱ABCD, if S~quadrilateral ABCD~ = 6.0, then the area of ​​the shaded part in the figure is (). Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "3 \n",
            "Response:<REASONING>\n",
            "We are given a parallelogram ABCD with area 6.0, and point E is any point inside it. The shaded regions are triangles ABE and CDE.\n",
            "\n",
            "Let’s denote the area of parallelogram ABCD as S = 6.0.\n",
            "\n",
            "The diagonals of a parallelogram bisect each other, but here we are not necessarily using diagonals — we are using an arbitrary point E inside the parallelogram.\n",
            "\n",
            "We can use the property that for any point E inside a parallelogram ABCD, the sum of the areas of triangles ABE and CDE equals half the area of the parallelogram, and the sum of the areas of triangles ADE and BCE also equals half the area of the parallelogram.\n",
            "\n",
            "But in this case, we are only asked for the shaded area, which consists of triangles ABE and CDE.\n",
            "\n",
            "Let’s denote:\n",
            "- Area of triangle ABE = x\n",
            "- Area of triangle CDE = y\n",
            "- Area of triangle ADE = p\n",
            "- Area of triangle BCE = q\n",
            "\n",
            "We know that:\n",
            "x + p + q + y = 6.0 (total area of parallelogram)\n",
            "\n",
            "Also, note that triangles ABE and CDE share the same height from E to AB and CD respectively? Not exactly, but we can use another approach.\n",
            "\n",
            "Consider that the area of triangle ABC is half the area of the parallelogram, since diagonal AC divides the parallelogram into two equal triangles.\n",
            "\n",
            "Similarly, area of triangle ADC is also half.\n",
            "\n",
            "But perhaps a better approach is to use the fact that for any point E inside parallelogram ABCD, the sum of the areas of triangles ABE and CDE is always half the area of the parallelogram.\n",
            "\n",
            "Let me verify this with a simple example.\n",
            "\n",
            "Suppose E is the intersection point of the diagonals. Then, the diagonals bisect each other, and each of the four triangles has area = 6.0 / 4 = 1.5. So ABE = 1.5, CDE = 1.5, total shaded = 3.0.\n",
            "\n",
            "Now, suppose E is at point A. Then triangle ABE has area 0 (since A and E coincide), and triangle CDE has area = area of triangle ACD = 3.0. So shaded area = 0 + 3.0 = 3.0.\n",
            "\n",
            "Suppose E is at point D. Then triangle ABE has area = area of triangle ABD = 3.0, and triangle CDE has area 0. So shaded area = 3.0 + 0 = 3.0.\n",
            "\n",
            "Suppose E is at point B. Then triangle ABE has area 0, and triangle CDE has area = area of triangle BCD = 3.0. Shaded = 3.0.\n",
            "\n",
            "Suppose E is at point C. Then triangle ABE has area = area of triangle ABC = 3.0, and triangle CDE has area 0. Shaded = 3.0.\n",
            "\n",
            "Suppose E is the midpoint of AD. Then triangle ABE: base AB, height from E to AB. Since E is midpoint of AD, and AD is a side, the height from E to AB is half the height from D to AB. So area of ABE = (1/2) * AB * (1/2 * h) = (1/2) * (1/2 * AB * h) = (1/2) * (area of triangle ABD) = (1/2) * (area of triangle ABD).\n",
            "\n",
            "Wait, area of triangle ABD = 3.0 (since diagonal BD divides parallelogram into two equal areas). So area of ABE = (1/2) * 3.0 = 1.5.\n",
            "\n",
            "Triangle CDE: base CD, height from E to CD. Since E is midpoint of AD, and CD is parallel to AB, the height from E to CD is the same as from E to AB? Not exactly. The height from E to CD is the distance from E to line CD. Since E is on AD, and CD is a side, the height from E to CD is the perpendicular distance.\n",
            "\n",
            "Actually, let's use coordinates to verify.\n",
            "\n",
            "Place parallelogram ABCD with A(0,0), B(2,0), C(3,2), D(1,2). Then area = base * height = 2 * 2 = 4.0? Wait, let's compute area using shoelace formula.\n",
            "\n",
            "Shoelace: A(0,0), B(2,0), C(3,2), D(1,2), back to A(0,0).\n",
            "\n",
            "Area = 1/2 | (0*0 + 2*2 + 3*2 + 1*0) - (0*2 + \n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>How many models in the figure achieve an Acc score greater than 60?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "2 \n",
            "Response:<REASONING>\n",
            "To determine how many models achieve an Acc score greater than 60, I need to examine the scatter plot and count the data points that lie above the horizontal line corresponding to Acc = 60.\n",
            "\n",
            "Step 1: Identify the horizontal line at Acc = 60.\n",
            "Step 2: Scan all the data points (models) on the plot and count those that are positioned above this line.\n",
            "Step 3: Note that the legend provides the model names, and each point is labeled with its corresponding model name.\n",
            "\n",
            "Let me go through the plot visually:\n",
            "- The point for GPT-4 (pink, top right) is clearly above 60.\n",
            "- The point for LLaMA-2-70B (purple, second from top right) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, slightly lower than the previous one) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower right) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also above 60.\n",
            "- The point for LLaMA-2-70B (purple, lower left) is also\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the age gap between these two people in image?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "28 \n",
            "Response:<REASONING>\n",
            "To determine the age gap between the two people in the image, I need to identify who they are and their birth years, or at least approximate their ages based on context and public records.\n",
            "\n",
            "The man in the image is clearly the former U.S. President Jimmy Carter. He was born on October 1, 1924.\n",
            "\n",
            "The woman in the image is Rosalynn Carter, his wife and former First Lady. She was born on August 18, 1928.\n",
            "\n",
            "Now, I calculate the age difference:\n",
            "\n",
            "Jimmy Carter’s birth year: 1924\n",
            "Rosalynn Carter’s birth year: 1928\n",
            "\n",
            "Age gap = |1924 - 1928| = 4 years.\n",
            "\n",
            "Since Jimmy Carter was born earlier, he is 4 years older than Rosalynn Carter.\n",
            "\n",
            "Therefore, the age gap is 4 years.\n",
            "\n",
            "</REASONING>\n",
            "\n",
            "<SOLUTION>4.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>How many people like the most preferred object in the whole chart?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "9 \n",
            "Response:<REASONING>\n",
            "To solve this question, I need to find the most preferred object across all categories in the chart. This means I need to identify the bar with the highest value on the x-axis (Number of People).\n",
            "\n",
            "Looking at the chart:\n",
            "- For \"junk\": the red bar (stove) is at ~3, and the blue bar (niece) is at ~9.\n",
            "- For \"horn\": the red bar (stove) is at ~7, and the blue bar (niece) is at ~5.\n",
            "- For \"meadow\": the red bar (stove) is at ~1, and the blue bar (niece) is at ~5.\n",
            "\n",
            "Now, I compare all these values: 3, 9, 7, 5, 1, and 5.\n",
            "\n",
            "The highest value is 9, which corresponds to the \"niece\" category for the \"junk\" object.\n",
            "\n",
            "Therefore, 9 people like the most preferred object in the whole chart.\n",
            "</REASONING>\n",
            "<SOLUTION>9.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the age gap between these two people in image?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "22 \n",
            "Response:<REASONING>\n",
            "To estimate the age gap between the two people in the image, I would need to make some assumptions since no explicit age information is provided. The image appears to be from a theatrical or film production, and the individuals are dressed in period costumes. The woman is wearing an elaborate gown, and the man is in a polka-dot costume, suggesting a historical or stylized setting.\n",
            "\n",
            "However, without knowing their actual birth years or ages, I cannot calculate a precise age gap. In such cases, the question might be expecting a hypothetical or illustrative answer based on typical age differences in such roles. For example, in theatrical productions, it’s common for a male lead to be older than the female lead, especially if the roles are romantic or dramatic. A common age gap in such pairings might be around 10 to 20 years.\n",
            "\n",
            "But since the question asks for a specific numerical answer and no data is provided, I must make a reasonable assumption. The most plausible answer, given the context of a theatrical scene and the typical age dynamics in such portrayals, would be a moderate gap, say 15 years. This is a common age difference in romantic or dramatic duets in historical settings.\n",
            "\n",
            "Alternatively, if we assume they are portraying characters of similar age (e.g., both in their 30s or 40s), the age gap might be small, perhaps 5 years. But without more context, the safest assumption for a \"typical\" age gap in such a scene is 15 years.\n",
            "\n",
            "Given the lack of specific data, I will provide a reasonable estimate based on common theatrical casting practices.\n",
            "\n",
            "</REASONING>\n",
            "\n",
            "<SOLUTION>15.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>How much more accurate is the most accurate algorithm compared the least accurate algorithm?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "5 \n",
            "Response:<REASONING>\n",
            "To solve this question, I need to:\n",
            "1. Identify the accuracy value for each algorithm from the bar chart.\n",
            "2. Determine which algorithm is the most accurate (highest bar).\n",
            "3. Determine which algorithm is the least accurate (lowest bar).\n",
            "4. Calculate the difference between the highest and lowest accuracy values.\n",
            "\n",
            "From the chart:\n",
            "- \"bronze\" has an accuracy of approximately 5.\n",
            "- \"spray\" has an accuracy of approximately 3.\n",
            "- \"cab\" has an accuracy of 8.\n",
            "\n",
            "The most accurate algorithm is \"cab\" with accuracy 8.\n",
            "The least accurate algorithm is \"spray\" with accuracy 3.\n",
            "\n",
            "The difference is: 8 - 3 = 5.\n",
            "\n",
            "Therefore, the most accurate algorithm is 5 units more accurate than the least accurate one.\n",
            "</REASONING>\n",
            "<SOLUTION>5.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Lines $l$, $m$, and $n$ are perpendicular bisectors of $\\triangle PQR$ and meet at $T$. If $TQ = 2x$, $PT = 3y - 1$, and $TR = 8$, find $z$.. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "3 \n",
            "Response:<REASONING>\n",
            "We are given that lines $l$, $m$, and $n$ are the perpendicular bisectors of triangle $PQR$ and they meet at point $T$. This means that $T$ is the circumcenter of triangle $PQR$. A key property of the circumcenter is that it is equidistant from all three vertices of the triangle. Therefore, we have:\n",
            "\n",
            "$PT = QT = RT$\n",
            "\n",
            "We are given:\n",
            "- $TQ = 2x$\n",
            "- $PT = 3y - 1$\n",
            "- $TR = 8$\n",
            "\n",
            "Since $T$ is the circumcenter, $PT = QT = RT$, so:\n",
            "\n",
            "$2x = 3y - 1 = 8$\n",
            "\n",
            "From $2x = 8$, we get $x = 4$.\n",
            "\n",
            "From $3y - 1 = 8$, we get $3y = 9$, so $y = 3$.\n",
            "\n",
            "Now, we look at the base of the triangle, segment $PR$. We are told that $PR = z + 4$ and that the perpendicular bisector $n$ meets $PR$ at its midpoint (since it is a perpendicular bisector). Also, the segment from $T$ to the midpoint of $PR$ is perpendicular to $PR$.\n",
            "\n",
            "But we are also given that the length of the segment from $T$ to the midpoint of $PR$ is 7 (the arrow labeled 7). This is the distance from $T$ to the midpoint of $PR$.\n",
            "\n",
            "In a triangle, the distance from the circumcenter $T$ to a side is not necessarily the same as the length of the side, but we can use the fact that the perpendicular bisector from $T$ to side $PR$ meets $PR$ at its midpoint, and the length of the segment from $T$ to that midpoint is 7.\n",
            "\n",
            "However, we can also use the Pythagorean theorem in triangle $TMR$ (or $TMP$) where $M$ is the midpoint of $PR$. Since $T$ is the circumcenter, and $M$ is the midpoint of $PR$, then $TM = 7$, and $PM = MR = \\frac{PR}{2} = \\frac{z + 4}{2}$.\n",
            "\n",
            "Also, since $TR = 8$ (which is the circumradius), and $TR$ is the hypotenuse of right triangle $TMR$, we have:\n",
            "\n",
            "$TR^2 = TM^2 + MR^2$\n",
            "\n",
            "Substituting the known values:\n",
            "\n",
            "$8^2 = 7^2 + \\left(\\frac{z + 4}{2}\\right)^2$\n",
            "\n",
            "$64 = 49 + \\left(\\frac{z + 4}{2}\\right)^2$\n",
            "\n",
            "$64 - 49 = \\left(\\frac{z + 4}{2}\\right)^2$\n",
            "\n",
            "$15 = \\left(\\frac{z + 4}{2}\\right)^2$\n",
            "\n",
            "Take square root of both sides:\n",
            "\n",
            "$\\sqrt{15} = \\frac{z + 4}{2}$ or $-\\sqrt{15} = \\frac{z + 4}{2}$\n",
            "\n",
            "Since lengths are positive, we take the positive root:\n",
            "\n",
            "$\\frac{z + 4}{2} = \\sqrt{15}$\n",
            "\n",
            "Multiply both sides by 2:\n",
            "\n",
            "$z + 4 = 2\\sqrt{15}$\n",
            "\n",
            "Then,\n",
            "\n",
            "$z = 2\\sqrt{15} - 4$\n",
            "\n",
            "But wait — let me check if this makes sense. The problem asks for $z$, and in the diagram, $z$ is part of a length label $z + 4$, which is the side $PR$. The value $2\\sqrt{15} - 4$ is approximately $2 \\times 3.873 - 4 = 7.746 - 4 = 3.746$, which is positive, so mathematically it's fine.\n",
            "\n",
            "However, let me double-check if I misinterpreted the diagram. The arrow labeled 7 is drawn from $T$ perpendicular to $PR$, and it meets $PR$ at its midpoint. So yes, $TM = 7$, and $TR = 8$ is the distance from $T$ to vertex $R$, which is the circumradius. So triangle $TMR$ is a right triangle with legs $TM = 7$ and $MR = \\frac{PR}{2} = \\frac{z + 4}{2}$, and hypotenuse $TR = 8$.\n",
            "\n",
            "So the equation is correct: $8^2 = 7^2 + \\left(\\frac{z + 4}{2}\\right)^2$.\n",
            "\n",
            "Therefore, $z = 2\\sqrt{15} - 4$.\n",
            "\n",
            "But the problem says \"\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>How many algorithms have accuracies higher than 9?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "0 \n",
            "Response:<REASONING>\n",
            "To solve this question, I need to analyze the bar chart titled \"Accuracy of different algorithms\".\n",
            "\n",
            "1. The chart displays two algorithms: \"icy\" and \"irony\".\n",
            "2. The y-axis represents \"Accuracy\", with values ranging from 0 to 10.\n",
            "3. I need to determine the accuracy for each algorithm by looking at the height of their respective bars:\n",
            "   - The bar for \"icy\" reaches the level of 2 on the y-axis. So, its accuracy is 2.\n",
            "   - The bar for \"irony\" reaches the level of 9 on the y-axis. So, its accuracy is 9.\n",
            "4. The question asks for the number of algorithms that have accuracies higher than 9.\n",
            "5. Comparing the accuracies:\n",
            "   - \"icy\": 2, which is not higher than 9.\n",
            "   - \"irony\": 9, which is not higher than 9 (it is equal to 9, not greater).\n",
            "6. Therefore, none of the algorithms have accuracies strictly higher than 9.\n",
            "\n",
            "The answer is 0.\n",
            "</REASONING>\n",
            "<SOLUTION>0.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the total value of the More bar?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "52 \n",
            "Response:<REASONING>\n",
            "The question asks for the total value of the \"More\" bar in the chart. Looking at the chart, the \"More\" category is represented by the dark brown bar. The value for the \"Total\" row under the \"More\" category is explicitly shown as 52. This is the value we need to report.\n",
            "\n",
            "There is no need for any calculation or aggregation since the value is directly provided in the chart for the \"Total\" row under the \"More\" category.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>52.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Find the perimeter of the parallelogram. Round to the nearest tenth if necessary.. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "44 \n",
            "Response:<REASONING>\n",
            "To find the perimeter of the parallelogram, we need the lengths of two adjacent sides. The diagram provides one side as 12 m, and the other side is labeled as 10 m. In a parallelogram, opposite sides are equal in length. Therefore, the two pairs of sides are 12 m and 10 m.\n",
            "\n",
            "The perimeter P of a parallelogram is given by the formula:\n",
            "P = 2 * (length + width)\n",
            "P = 2 * (12 m + 10 m)\n",
            "P = 2 * 22 m\n",
            "P = 44 m\n",
            "\n",
            "The given angle (60°) and the height (implied by the right angle symbol) are not needed to calculate the perimeter, since perimeter only depends on side lengths. However, we can verify the height if needed, but it's not required for this question.\n",
            "\n",
            "Therefore, the perimeter is 44.0 meters.\n",
            "</REASONING>\n",
            "<SOLUTION>44.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>如图，在△ABC中，点D是△ABC的内心，连接DB，DC，过点D作EF∥BC分别交AB、AC于点E、F，若BE+CF＝8，则EF的长度为（）. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "8 \n",
            "Response:<REASONING>\n",
            "We are given that D is the incenter of triangle ABC, meaning it is the point where the angle bisectors of the triangle meet, and it is equidistant from all three sides. The line EF is drawn through D parallel to BC, intersecting AB at E and AC at F.\n",
            "\n",
            "Since D is the incenter, the angle bisectors of ∠ABC and ∠ACB pass through D. Because EF is parallel to BC, we can use properties of similar triangles and angle bisectors.\n",
            "\n",
            "Consider triangle ABC with incenter D. Since EF ∥ BC, triangles AEF and ABC are similar.\n",
            "\n",
            "Now, let’s consider the segments BE and CF. We are given that BE + CF = 8.\n",
            "\n",
            "We need to find the length of EF.\n",
            "\n",
            "Key insight: In a triangle with incenter D and a line EF through D parallel to BC, the segments BE and CF can be related to the segments AE and AF. However, a more direct and elegant approach uses the property that when a line is drawn through the incenter parallel to the base, it creates a relationship with the sum of the segments on the sides.\n",
            "\n",
            "Let’s denote the incenter as D. Since D is the incenter, it lies on the angle bisectors of ∠B and ∠C. Therefore, ∠EBD = ∠DBC and ∠FCD = ∠DCB.\n",
            "\n",
            "Because EF ∥ BC, we have ∠EDB = ∠DBC (alternate interior angles) and ∠FDC = ∠DCB (alternate interior angles).\n",
            "\n",
            "Therefore, ∠EBD = ∠EDB and ∠FCD = ∠FDC.\n",
            "\n",
            "This implies that triangles EBD and FDC are isosceles:\n",
            "\n",
            "- In triangle EBD: ∠EBD = ∠EDB ⇒ BE = ED\n",
            "- In triangle FDC: ∠FCD = ∠FDC ⇒ CF = FD\n",
            "\n",
            "Therefore, BE = ED and CF = FD.\n",
            "\n",
            "Now, EF = ED + FD = BE + CF.\n",
            "\n",
            "Given that BE + CF = 8, it follows that EF = 8.\n",
            "\n",
            "This is a known property: in a triangle, if a line is drawn through the incenter parallel to the base, then the length of that line segment equals the sum of the segments cut off on the other two sides.\n",
            "\n",
            "Therefore, EF = BE + CF = 8.\n",
            "\n",
            "</REASONING>\n",
            "\n",
            "<SOLUTION>8.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>How much money does Damon need to buy a grilled steak and a mushroom pizza? (Unit: $). Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "24 \n",
            "Response:<REASONING>\n",
            "To determine how much money Damon needs to buy a grilled steak and a mushroom pizza, I need to:\n",
            "1. Identify the price of the grilled steak from the menu.\n",
            "2. Identify the price of the mushroom pizza from the menu.\n",
            "3. Add these two prices together to get the total cost.\n",
            "\n",
            "From the menu:\n",
            "- Grilled steak costs $13.\n",
            "- Mushroom pizza costs $11.\n",
            "\n",
            "Adding them: $13 + $11 = $24.\n",
            "\n",
            "Therefore, Damon needs $24 to buy both items.\n",
            "</REASONING>\n",
            "<SOLUTION>\n",
            "24.0\n",
            "</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>For how many years that the percentage value over 4?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "6 \n",
            "Response:<REASONING>\n",
            "To determine for how many years the percentage value was over 4%, I need to examine the line chart and identify each year where the data point is above the 4% mark on the y-axis.\n",
            "\n",
            "Looking at the data points year by year:\n",
            "- 2005: 4.2% → not over 4% (it's equal to 4%)\n",
            "- 2006: 4.3% → over 4%\n",
            "- 2007: 4.8% → over 4%\n",
            "- 2008: 4.9% → over 4%\n",
            "- 2009: 5.3% → over 4%\n",
            "- 2010: 4.1% → not over 4%\n",
            "- 2011: 3.8% → not over 4%\n",
            "- 2012: 3.8% → not over 4%\n",
            "- 2013: 3.4% → not over 4%\n",
            "- 2014: 3.6% → not over 4%\n",
            "- 2015: 3.6% → not over 4%\n",
            "- 2016: 3.5% → not over 4%\n",
            "- 2017: 3.5% → not over 4%\n",
            "- 2018*: 3.7% → not over 4%\n",
            "- 2019**: 3.5% → not over 4%\n",
            "\n",
            "The years where the percentage is over 4% are: 2006, 2007, 2008, and 2009.\n",
            "\n",
            "Counting these, we have 4 years.\n",
            "\n",
            "Note: The problem asks for \"over 4%\", so values equal to 4% (like 2005) are not counted. The values 4.1% and above are counted, but 4.0% is not.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>4.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Find $m \\angle N C L$. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "120 \n",
            "Response:<REASONING>\n",
            "We are given a circle with center C, and three radii: CR, CM, and CL. The angles at the center are labeled: angle RCM is (x - 1)°, angle MCL is 60°, and angle LCR is (3x + 5)°. However, looking at the diagram, the angle labeled (3x + 5)° is between radii CR and CL, so it should be angle RCL. But the angle between CR and CM is labeled (x - 1)°, and between CM and CL is 60°.\n",
            "\n",
            "The key is that the three angles RCM, MCL, and LCR together form the full circle around center C, which is 360°. So:\n",
            "\n",
            "angle RCM + angle MCL + angle LCR = 360°\n",
            "\n",
            "Substituting the given expressions:\n",
            "\n",
            "(x - 1)° + 60° + (3x + 5)° = 360°\n",
            "\n",
            "Simplify the left side:\n",
            "\n",
            "x - 1 + 60 + 3x + 5 = 360\n",
            "\n",
            "Combine like terms:\n",
            "\n",
            "( x + 3x ) + ( -1 + 60 + 5 ) = 360\n",
            "\n",
            "4x + 64 = 360\n",
            "\n",
            "Now solve for x:\n",
            "\n",
            "4x = 360 - 64\n",
            "\n",
            "4x = 296\n",
            "\n",
            "x = 296 / 4\n",
            "\n",
            "x = 74\n",
            "\n",
            "Now, we are asked to find m∠NCL. Looking at the diagram, point N is on the arc between M and L, and angle MCL is given as 60°. Since angle MCL is the angle between radii CM and CL, and point N is on the arc ML, then angle NCL is the same as angle MCL if N is on the arc ML, which is the case. So angle NCL is the angle between radii CN and CL. But since angle MCL is 60°, and point N is on the arc ML, then angle NCL is the same as angle MCL, which is 60°.\n",
            "\n",
            "Wait, let me re-examine. The diagram shows points R, M, N, L around the circle. The angle labeled 60° is between CM and CN, not CM and CL. The angle labeled (3x + 5)° is between CR and CL. The angle between CM and CL is not labeled, but we can find it.\n",
            "\n",
            "Actually, let's look carefully. The three central angles are: angle RCM, angle MCN, and angle NCL. But the diagram labels angle RCM as (x - 1)°, angle MCN as 60°, and angle NCL as (3x + 5)°? Wait, that doesn't make sense because (3x + 5)° is drawn between CR and CL, not between NC and CL.\n",
            "\n",
            "Looking at the diagram again: the angle (3x + 5)° is between CR and CL. The angle (x - 1)° is between CR and CM. The angle 60° is between CM and CN. But then point N is between M and L, so the angle between CM and CL should be angle MCN + angle NCL.\n",
            "\n",
            "So the total central angle from CR to CL is angle RCM + angle MCN + angle NCL.\n",
            "\n",
            "So:\n",
            "\n",
            "(x - 1) + 60 + (3x + 5) = 360\n",
            "\n",
            "Which is what I had before.\n",
            "\n",
            "So x = 74.\n",
            "\n",
            "Now, we need to find m∠NCL. But in the diagram, the angle labeled (3x + 5)° is between CR and CL, not between NC and CL. So that can't be angle NCL.\n",
            "\n",
            "Perhaps the labeling is a bit confusing. Let me read the problem again: \"Find m∠NCL\".\n",
            "\n",
            "Looking at the diagram, point N is on the arc between M and L. So angle NCL is the angle at center C between points N, C, and L. So it's the angle between radii CN and CL.\n",
            "\n",
            "The given angles are:\n",
            "\n",
            "- Angle between CR and CM: (x - 1)°\n",
            "- Angle between CM and CN: 60° (labeled)\n",
            "- Angle between CR and CL: (3x + 5)°\n",
            "\n",
            "So, the angle between CN and CL is not directly given. But we can find it by subtracting the angle between CN and CM from the angle between CL and CM.\n",
            "\n",
            "First, let's find the angle between CM and CL.\n",
            "\n",
            "From the diagram, the angle between CM and CL is the sum of angle MCN and angle NCL. But we don't know angle NCL.\n",
            "\n",
            "Alternatively, the total angle from\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Subtract all tiny purple shiny cubes. Subtract all large purple balls. How many objects are left?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "9 \n",
            "Response:<REASONING>\n",
            "First, I need to identify all the objects in the image.\n",
            "\n",
            "Looking at the image, I can see:\n",
            "- 1 small gray cylinder (left side, behind the teal sphere)\n",
            "- 1 teal sphere (left side)\n",
            "- 1 large green cylinder (front center)\n",
            "- 1 blue shiny cylinder (left side, behind the teal sphere)\n",
            "- 1 red shiny cylinder (center)\n",
            "- 1 small red cylinder (behind the red shiny cylinder)\n",
            "- 1 purple sphere (center back)\n",
            "- 1 yellow cube (right side)\n",
            "- 1 gold shiny sphere (right side, in front of the yellow cube)\n",
            "- 1 small blue sphere (center front, in front of the green cylinder)\n",
            "\n",
            "That's a total of 10 objects.\n",
            "\n",
            "Now, the question asks to:\n",
            "1. Subtract all tiny purple shiny cubes.\n",
            "2. Subtract all large purple balls.\n",
            "\n",
            "Let's check if any of these conditions are met.\n",
            "\n",
            "First, are there any \"tiny purple shiny cubes\"? Looking at the image, there is a yellow cube, but it's not purple. There are no purple cubes at all, let alone tiny and shiny ones. So, we subtract 0 objects for this condition.\n",
            "\n",
            "Second, are there any \"large purple balls\"? There is one purple sphere in the background. It is large compared to the other objects (like the small blue sphere), and it is purple. So, we need to subtract this one object.\n",
            "\n",
            "Therefore, we start with 10 objects, subtract 1 (the large purple ball), and subtract 0 (no tiny purple shiny cubes).\n",
            "\n",
            "So, 10 - 1 = 9 objects left.\n",
            "\n",
            "</REASONING>\n",
            "\n",
            "<SOLUTION>9.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Subtract all brown things. Subtract all tiny blue metallic objects. How many objects are left?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "2 \n",
            "Response:<REASONING>\n",
            "We are given an image with four objects:\n",
            "1. A large red metallic sphere.\n",
            "2. A gold (brownish) metallic cylinder.\n",
            "3. A small blue matte cube.\n",
            "4. A small blue metallic sphere.\n",
            "\n",
            "The instruction is to:\n",
            "- Subtract all brown things. The gold cylinder is the only brown thing (or brownish/golden object). So we remove that.\n",
            "- Subtract all tiny blue metallic objects. The small blue metallic sphere fits this description. So we remove that.\n",
            "\n",
            "After these subtractions, we are left with:\n",
            "- The large red metallic sphere.\n",
            "- The small blue matte cube.\n",
            "\n",
            "Therefore, 2 objects remain.\n",
            "\n",
            "Note: The cube is not metallic, so it is not removed by the second instruction. The sphere is tiny and metallic, so it is removed. The cylinder is brown, so it is removed.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>2.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>what is the total volume of the measuring cup?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "1000 \n",
            "Response:<REASONING>\n",
            "To determine the total volume of the measuring cup, I would examine the markings on the side of the cup. Measuring cups are typically calibrated in standard units like milliliters (mL) or liters (L), and often also include markings for other common ingredients like sugar, flour, or milk, but the volume scale is the primary indicator.\n",
            "\n",
            "In the image, the measuring cup has markings in milliliters (mL) and liters (L). The highest marking visible is “1000 mL” (which is equivalent to 1 liter). The cup also has markings for 500 mL, 250 mL, 100 mL, 50 mL, 20 mL, 10 mL, and even smaller increments like 5 mL or 2 mL. The presence of the “1000 mL” mark at the top indicates that the maximum capacity of the cup is 1000 mL.\n",
            "\n",
            "Additionally, the text on the cup includes “1000 g” which suggests it’s also calibrated for weight, likely for ingredients like flour or sugar, but the volume scale is still in mL.\n",
            "\n",
            "Therefore, the total volume of the measuring cup is 1000 mL.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>1000.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>As shown in the figure, line segment AB = 10.0, M is the midpoint of line segment AB, C is the midpoint of line segment MB, N is a point of line segment AM, and MN = 1.0, the length of line segment NC (). Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "3.5 \n",
            "Response:<REASONING>\n",
            "We are given:\n",
            "- AB = 10.0\n",
            "- M is the midpoint of AB, so AM = MB = 5.0\n",
            "- C is the midpoint of MB, so MC = CB = 2.5\n",
            "- N is a point on AM such that MN = 1.0\n",
            "\n",
            "We need to find the length of NC.\n",
            "\n",
            "Step 1: Determine the position of N.\n",
            "Since N is on AM and MN = 1.0, and AM = 5.0, then AN = AM - MN = 5.0 - 1.0 = 4.0.\n",
            "\n",
            "So, point N is located 4.0 units from A along AB.\n",
            "\n",
            "Step 2: Determine the position of C.\n",
            "C is the midpoint of MB. Since MB = 5.0, and M is at 5.0 from A (since AM = 5.0), then C is 2.5 units from M towards B. So, the position of C from A is AM + MC = 5.0 + 2.5 = 7.5.\n",
            "\n",
            "Step 3: Calculate NC.\n",
            "N is at 4.0 from A, and C is at 7.5 from A. So, the distance NC = |7.5 - 4.0| = 3.5.\n",
            "\n",
            "Alternatively, we can think of it as:\n",
            "NC = MC + MN = 2.5 + 1.0 = 3.5.\n",
            "\n",
            "But wait, let's verify this. If N is at 4.0 from A, and C is at 7.5 from A, then NC = 7.5 - 4.0 = 3.5. This is correct.\n",
            "\n",
            "Another way: Since M is at 5.0 from A, and N is at 4.0 from A, then MN = 1.0 (given). C is at 7.5 from A. So from N to C is 7.5 - 4.0 = 3.5.\n",
            "\n",
            "So, the length of NC is 3.5.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>3.5</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Colton wants to buy 1+3/10 kilograms of English muffins. How much will he spend? (Unit: $). Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "10.4 \n",
            "Response:<REASONING>\n",
            "To find out how much Colton will spend on 1+3/10 kilograms of English muffins, we need to:\n",
            "\n",
            "1. Determine the price per kilogram of English muffins from the given table.\n",
            "   - From the table, English muffins cost $8 per kilogram.\n",
            "\n",
            "2. Convert the quantity Colton wants to buy into an improper fraction or decimal for easier calculation.\n",
            "   - 1+3/10 kilograms = 1.3 kilograms (since 3/10 = 0.3).\n",
            "\n",
            "3. Multiply the price per kilogram by the quantity in kilograms.\n",
            "   - Cost = Price per kg × Quantity in kg\n",
            "   - Cost = $8 × 1.3\n",
            "\n",
            "4. Perform the multiplication.\n",
            "   - $8 × 1.3 = $10.40\n",
            "\n",
            "Therefore, Colton will spend $10.40.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>\n",
            "10.4</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Find $m \\angle K$. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "100 \n",
            "Response:<REASONING>\n",
            "We are given a quadrilateral JKLM with the following information:\n",
            "- Sides JM and KL are both labeled 6, so JM = KL = 6.\n",
            "- Angle at M, ∠JML (or ∠M), is 80°.\n",
            "- The figure appears to be a trapezoid with JM and KL as the non-parallel sides (legs), and JK and ML as the bases (top and bottom). However, since the top side JK is not labeled, and the bottom side ML is not labeled, we need to deduce the type of quadrilateral.\n",
            "\n",
            "But note: the figure has two sides labeled 6 and an angle of 80° at M. The arrows on JK and ML suggest that JK and ML are parallel (as in a trapezoid or parallelogram). However, since JM and KL are equal and non-adjacent, and the figure is symmetric, it is likely an isosceles trapezoid.\n",
            "\n",
            "In an isosceles trapezoid, the base angles are equal. That is, ∠M and ∠L are the base angles at the bottom, and ∠J and ∠K are the base angles at the top.\n",
            "\n",
            "But wait — if JM and KL are the legs (non-parallel sides), and both are 6, and if the top and bottom bases are parallel, then it is an isosceles trapezoid.\n",
            "\n",
            "In an isosceles trapezoid, the base angles are equal. So ∠M = ∠L, and ∠J = ∠K.\n",
            "\n",
            "We are given ∠M = 80°, so ∠L = 80°.\n",
            "\n",
            "The sum of the interior angles of any quadrilateral is 360°.\n",
            "\n",
            "So, ∠M + ∠L + ∠J + ∠K = 360°.\n",
            "\n",
            "Substituting known values:\n",
            "\n",
            "80° + 80° + ∠J + ∠K = 360°\n",
            "\n",
            "160° + ∠J + ∠K = 360°\n",
            "\n",
            "∠J + ∠K = 200°\n",
            "\n",
            "But since it's an isosceles trapezoid, ∠J = ∠K.\n",
            "\n",
            "Therefore, 2∠K = 200°\n",
            "\n",
            "∠K = 100°\n",
            "\n",
            "Alternatively, we can think geometrically: if the bottom base angles are 80°, then the top base angles are supplementary to the bottom base angles if the legs are slanting inward — but in a trapezoid, consecutive angles between the bases are supplementary.\n",
            "\n",
            "Wait — that’s a key point.\n",
            "\n",
            "In a trapezoid with parallel bases, consecutive angles between the bases are supplementary. That is, ∠M + ∠J = 180°, and ∠L + ∠K = 180°.\n",
            "\n",
            "This is because the legs are transversals cutting parallel lines.\n",
            "\n",
            "So, since ∠M = 80°, then ∠J = 180° - 80° = 100°.\n",
            "\n",
            "Similarly, ∠L = 80°, so ∠K = 180° - 80° = 100°.\n",
            "\n",
            "But wait — is this trapezoid isosceles? The legs are both 6, so yes, it is isosceles, and in an isosceles trapezoid, the base angles are equal, so ∠M = ∠L = 80°, and the top angles ∠J = ∠K = 100°.\n",
            "\n",
            "So both methods give ∠K = 100°.\n",
            "\n",
            "Therefore, the measure of angle K is 100 degrees.\n",
            "\n",
            "</REASONING>\n",
            "\n",
            "<SOLUTION>\n",
            "100.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Phenylalanine (Phe, 5) is a naturally occurring amino acid. What is the energy of interaction between its phenyl group and the electric dipole moment of a neighbouring peptide group? Take the distance between the groups as $4.0 \\mathrm{~nm}$ and treat the phenyl group as a benzene molecule. The magnitude of the dipole moment of the peptide group is $\\mu=1.3 \\mathrm{D}$ and the polarizability volume of benzene is $\\alpha^{\\prime}=1.04 \\times 10^{-29} \\mathrm{~m}^3$.\n",
            ". Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "-4.3 \n",
            "Response:<REASONING>\n",
            "To calculate the energy of interaction between the phenyl group (modeled as benzene) and the electric dipole moment of a neighboring peptide group, we can use the formula for the interaction energy between a dipole and an induced dipole:\n",
            "\n",
            "E = - (μ² α' ) / (4πε₀ r⁴)\n",
            "\n",
            "where:\n",
            "- μ is the dipole moment of the peptide group (1.3 D),\n",
            "- α' is the polarizability volume of benzene (1.04 × 10⁻²⁹ m³),\n",
            "- r is the distance between the groups (4.0 nm),\n",
            "- ε₀ is the permittivity of free space (8.854 × 10⁻¹² F/m),\n",
            "- and we must convert dipole moment from Debye (D) to SI units (C·m).\n",
            "\n",
            "Conversion factor: 1 D = 3.33564 × 10⁻¹⁰ C·m.\n",
            "\n",
            "So, μ = 1.3 × 3.33564 × 10⁻¹⁰ C·m = 4.336332 × 10⁻¹⁰ C·m.\n",
            "\n",
            "Distance r = 4.0 nm = 4.0 × 10⁻⁹ m.\n",
            "\n",
            "We will plug these values into the formula.\n",
            "\n",
            "First, compute μ²:\n",
            "μ² = (4.336332 × 10⁻¹⁰)² = 1.8803 × 10⁻¹⁸ C²·m²\n",
            "\n",
            "Then compute r⁴:\n",
            "r⁴ = (4.0 × 10⁻⁹)⁴ = 2.56 × 10⁻²⁸ m⁴\n",
            "\n",
            "Now compute the constant factor:\n",
            "4πε₀ = 4 × π × 8.854 × 10⁻¹² = 1.11265 × 10⁻¹₀ F/m (but we'll keep it as 4πε₀ for calculation).\n",
            "\n",
            "Now compute E:\n",
            "E = - (μ² α') / (4πε₀ r⁴)\n",
            "\n",
            "Plug in the numbers:\n",
            "Numerator: μ² α' = (1.8803 × 10⁻¹⁸) × (1.04 × 10⁻²⁹) = 1.9555 × 10⁻⁴⁷\n",
            "\n",
            "Denominator: 4πε₀ r⁴ = (1.11265 × 10⁻¹⁰) × (2.56 × 10⁻²⁸) = 2.844 × 10⁻³⁷\n",
            "\n",
            "So E = - (1.9555 × 10⁻⁴⁷) / (2.844 × 10⁻³⁷) = -6.873 × 10⁻¹¹ J\n",
            "\n",
            "We can also express this in more convenient units (e.g., kJ/mol) for biological contexts, but the question asks for the energy in joules, so we'll leave it as is.\n",
            "\n",
            "However, let's double-check the formula. The standard formula for dipole-induced dipole interaction is:\n",
            "\n",
            "E = - (μ² α') / (4πε₀ r⁴)\n",
            "\n",
            "This is correct for a point dipole and a polarizable molecule.\n",
            "\n",
            "Let me recalculate with more precision.\n",
            "\n",
            "First, μ = 1.3 D = 1.3 × 3.33564 × 10⁻¹⁰ = 4.336332 × 10⁻¹⁰ C·m\n",
            "\n",
            "μ² = (4.336332 × 10⁻¹⁰)² = 1.8803 × 10⁻¹⁸\n",
            "\n",
            "α' = 1.04 × 10⁻²⁹ m³\n",
            "\n",
            "μ² α' = 1.8803 × 10⁻¹⁸ × 1.04 × 10⁻²⁹ = 1.9555 × 10⁻⁴⁷\n",
            "\n",
            "r = 4.0 × 10⁻⁹ m\n",
            "\n",
            "r⁴ = (4.0 × 10⁻⁹)⁴ = 256 × 10⁻³⁶ = 2.56 × 10⁻³⁴? Wait, no:\n",
            "\n",
            "(4.0 × 10⁻⁹)² = 16 × 10⁻¹⁸ = 1.6 × 10⁻¹⁷\n",
            "\n",
            "(4.0 × 10⁻⁹)⁴\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the perimeter of the shape?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "9 \n",
            "Response:<REASONING>\n",
            "The shape shown is an equilateral triangle, as all three sides are labeled as 3 yards each. The perimeter of any polygon is the sum of the lengths of its sides. Therefore, to find the perimeter, we add the lengths of all three sides: 3 yd + 3 yd + 3 yd.\n",
            "\n",
            "Perimeter = 3 + 3 + 3 = 9 yd.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>9.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Some friends compared the sizes of their stuffed animal collections. What is the median of the numbers?'. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "9 \n",
            "Response:<REASONING>\n",
            "To find the median of a set of numbers, I need to follow these steps:\n",
            "\n",
            "1. List all the numbers from the table: 9, 9, 8, 6, 10.\n",
            "2. Arrange them in ascending order: 6, 8, 9, 9, 10.\n",
            "3. Count the total number of values. There are 5 values.\n",
            "4. Since the number of values is odd (5), the median is the middle value. In a list of 5 numbers, the middle position is the 3rd one (since (5+1)/2 = 3).\n",
            "5. The 3rd number in the ordered list [6, 8, 9, 9, 10] is 9.\n",
            "\n",
            "Therefore, the median is 9.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>9.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>what number does the smaller arrow point to?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "1020 \n",
            "Response:<REASONING>\n",
            "The object in the image is a barometer, specifically a vintage-style aneroid barometer with a circular dial. The dial has markings for atmospheric pressure, with labels such as \"Regn\" (for rain), \"Normal\" (for normal conditions), and \"Sol\" (for sun). The smaller arrow (the shorter hand) points to a specific pressure level on the dial. \n",
            "\n",
            "Looking closely at the dial:\n",
            "- The smaller arrow is pointing to a position between \"Normal\" and \"Regn\".\n",
            "- The scale is marked with numbers from 0 to 10, with \"Normal\" positioned at approximately 7.\n",
            "- The smaller arrow is pointing to a value just below 7, specifically at the 6.5 mark (halfway between 6 and 7).\n",
            "\n",
            "Therefore, the smaller arrow points to 6.5.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>6.5</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Subtract all large rubber spheres. Subtract all big shiny cylinders. How many objects are left?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "4 \n",
            "Response:<REASONING>\n",
            "We are given a scene with 6 objects:\n",
            "1. A large green sphere (rubber-like texture)\n",
            "2. A large purple cylinder (rubber-like texture)\n",
            "3. A large yellow cube (rubber-like texture)\n",
            "4. A large shiny silver cube (metallic)\n",
            "5. A large shiny yellow cylinder (metallic)\n",
            "6. A small purple sphere (rubber-like texture)\n",
            "\n",
            "The instruction is to:\n",
            "- Subtract all large rubber spheres.\n",
            "  - The large green sphere is a large rubber sphere. So we remove it.\n",
            "- Subtract all big shiny cylinders.\n",
            "  - The large shiny yellow cylinder is a big shiny cylinder. So we remove it.\n",
            "\n",
            "Objects removed: 2 (the green sphere and the yellow cylinder).\n",
            "\n",
            "Objects remaining: 4 (the purple cylinder, the yellow cube, the silver cube, and the small purple sphere).\n",
            "\n",
            "Therefore, 4 objects are left.\n",
            "</REASONING>\n",
            "<SOLUTION>4.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>One of the most dramatic videos on the web (but entirely fictitious) supposedly shows a man sliding along a long water slide and then being launched into the air to land in a water pool. Let's attach some reasonable numbers to such a flight to calculate the velocity with which the man would have hit the water. Figure indicates the launch and landing sites and includes a superimposed coordinate system with its origin conveniently located at the launch site. From the video we take the horizontal flight distance as $D=20.0 \\mathrm{~m}$, the flight time as $t=2.50 \\mathrm{~s}$, and the launch angle as $\\theta_0=40.0^{\\circ}$. Find the magnitude of the velocity at launch and at landing.. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "10.44 \n",
            "Response:<REASONING>\n",
            "To solve this problem, we need to find the magnitude of the velocity at launch and at landing for a projectile motion scenario.\n",
            "\n",
            "Given:\n",
            "- Horizontal distance: D = 20.0 m\n",
            "- Flight time: t = 2.50 s\n",
            "- Launch angle: θ₀ = 40.0°\n",
            "\n",
            "Step 1: Find the horizontal component of initial velocity (v₀ₓ) using the horizontal motion equation.\n",
            "Since there is no horizontal acceleration (air resistance neglected), horizontal velocity is constant:\n",
            "v₀ₓ = D / t = 20.0 m / 2.50 s = 8.00 m/s\n",
            "\n",
            "Step 2: Find the initial vertical component of velocity (v₀ᵧ) using the launch angle.\n",
            "v₀ᵧ = v₀ₓ * tan(θ₀) = 8.00 m/s * tan(40.0°)\n",
            "\n",
            "But wait — this is incorrect reasoning. Actually, we should use the fact that the horizontal component is v₀ₓ = v₀ * cos(θ₀), so we can solve for the initial speed v₀.\n",
            "\n",
            "From v₀ₓ = v₀ * cos(θ₀)\n",
            "=> v₀ = v₀ₓ / cos(θ₀) = 8.00 m/s / cos(40.0°)\n",
            "\n",
            "Let me calculate that.\n",
            "\n",
            "cos(40.0°) ≈ 0.7660\n",
            "v₀ = 8.00 / 0.7660 ≈ 10.415 m/s\n",
            "\n",
            "So initial velocity magnitude is approximately 10.415 m/s.\n",
            "\n",
            "Step 3: Find the vertical component of initial velocity.\n",
            "v₀ᵧ = v₀ * sin(θ₀) = 10.415 * sin(40.0°)\n",
            "\n",
            "sin(40.0°) ≈ 0.6428\n",
            "v₀ᵧ ≈ 10.415 * 0.6428 ≈ 6.700 m/s\n",
            "\n",
            "Step 4: Find the vertical component of velocity at landing.\n",
            "Use the kinematic equation for vertical motion:\n",
            "vᵧ = v₀ᵧ - g*t\n",
            "where g = 9.81 m/s² (standard gravity)\n",
            "\n",
            "vᵧ = 6.700 - (9.81 * 2.50) = 6.700 - 24.525 = -17.825 m/s\n",
            "\n",
            "The negative sign indicates downward direction.\n",
            "\n",
            "Step 5: Find the magnitude of velocity at landing.\n",
            "The velocity at landing has two components: horizontal (vₓ) and vertical (vᵧ).\n",
            "Since horizontal velocity is constant, vₓ = v₀ₓ = 8.00 m/s (same as initial horizontal component).\n",
            "\n",
            "Magnitude of velocity at landing = √(vₓ² + vᵧ²) = √(8.00² + (-17.825)²) = √(64.00 + 317.90) = √381.90 ≈ 19.54 m/s\n",
            "\n",
            "Step 6: Find the magnitude of velocity at launch.\n",
            "That's simply the initial speed v₀ = 10.415 m/s.\n",
            "\n",
            "So, to summarize:\n",
            "- Initial velocity magnitude: 10.415 m/s\n",
            "- Landing velocity magnitude: 19.54 m/s\n",
            "\n",
            "Let me verify the calculations.\n",
            "\n",
            "v₀ₓ = 20.0 / 2.50 = 8.00 m/s — correct.\n",
            "\n",
            "v₀ = v₀ₓ / cos(40.0°) = 8.00 / 0.7660 ≈ 10.415 m/s — correct.\n",
            "\n",
            "v₀ᵧ = 10.415 * sin(40.0°) = 10.415 * 0.6428 ≈ 6.700 m/s — correct.\n",
            "\n",
            "vᵧ = 6.700 - 9.81*2.50 = 6.700 - 24.525 = -17.825 m/s — correct.\n",
            "\n",
            "v_land = √(8.00² + 17.825²) = √(64 + 317.90) = √381.90 ≈ 19.54 m/s — correct.\n",
            "\n",
            "Therefore, the initial velocity is approximately 10.42 m/s, and the landing velocity is approximately 19.54 m/s.\n",
            "\n",
            "Note: The problem asks for \"the magnitude of the velocity at launch and at landing\". It does not specify which one first. Since the\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>$a=14, b=48,$ and $c=50$ find $cosA$. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "0.96 \n",
            "Response:<REASONING>\n",
            "We are given a right triangle ABC with the right angle at C. The sides are labeled as follows:\n",
            "- Side a (opposite angle A) is BC = 14\n",
            "- Side b (opposite angle B) is AC = 48\n",
            "- Side c (hypotenuse, opposite the right angle C) is AB = 50\n",
            "\n",
            "We are to find cos(A), which is the cosine of angle A.\n",
            "\n",
            "In a right triangle, the cosine of an angle is defined as the ratio of the length of the adjacent side to the hypotenuse.\n",
            "\n",
            "For angle A:\n",
            "- The adjacent side is AC, which has length b = 48\n",
            "- The hypotenuse is AB, which has length c = 50\n",
            "\n",
            "Therefore:\n",
            "cos(A) = adjacent / hypotenuse = b / c = 48 / 50\n",
            "\n",
            "We can simplify this fraction:\n",
            "48 / 50 = 24 / 25\n",
            "\n",
            "As a decimal, 24 / 25 = 0.96\n",
            "\n",
            "We can verify this using the Pythagorean theorem to confirm the triangle is right-angled:\n",
            "a² + b² = 14² + 48² = 196 + 2304 = 2500\n",
            "c² = 50² = 2500\n",
            "Since a² + b² = c², the triangle is indeed right-angled at C, and our setup is correct.\n",
            "\n",
            "Thus, cos(A) = 48 / 50 = 0.96\n",
            "</REASONING>\n",
            "\n",
            "<SOLUTION>\n",
            "0.96\n",
            "</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>How many people prefer the most preferred object?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "7 \n",
            "Response:<REASONING>\n",
            "To determine how many people prefer the most preferred object, I need to:\n",
            "\n",
            "1. Examine the bar chart titled \"Most preferred objects\".\n",
            "2. Identify the two categories on the y-axis: \"pupil\" and \"nurse\".\n",
            "3. Read the corresponding values on the x-axis (\"Number of People\") for each category.\n",
            "   - The bar for \"pupil\" extends to approximately 7 on the x-axis.\n",
            "   - The bar for \"nurse\" extends to exactly 6 on the x-axis.\n",
            "4. Determine which object is the most preferred: the one with the highest number of people. Since 7 > 6, \"pupil\" is the most preferred object.\n",
            "5. The value corresponding to \"pupil\" is 7.\n",
            "\n",
            "Therefore, 7 people prefer the most preferred object.\n",
            "</REASONING>\n",
            "\n",
            "<SOLUTION>7.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the sum of accuracies of the algorithm liver for all the datasets?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "24 \n",
            "Response:<REASONING>\n",
            "To find the sum of accuracies of the algorithm \"liver\" for all the datasets, I need to:\n",
            "1. Identify the accuracy values for \"liver\" across all five datasets (desert, debt, bee, saw, wool) from the bar chart.\n",
            "2. Add these values together.\n",
            "\n",
            "From the chart:\n",
            "- For \"liver\" (x-axis label):\n",
            "  - desert (green): 2\n",
            "  - debt (purple): 5\n",
            "  - bee (blue): 3\n",
            "  - saw (brown): 7\n",
            "  - wool (light blue): 7\n",
            "\n",
            "Sum = 2 + 5 + 3 + 7 + 7 = 24\n",
            "\n",
            "Therefore, the sum of accuracies for the algorithm \"liver\" is 24.\n",
            "</REASONING>\n",
            "<SOLUTION>24.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Subtract all large yellow matte cubes. Subtract all metal things. How many objects are left?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "2 \n",
            "Response:<REASONING>\n",
            "Let's break down the problem step by step.\n",
            "\n",
            "1. First, identify all objects in the image:\n",
            "   - There is a small gold cylinder (metallic, shiny).\n",
            "   - There is a small red sphere (metallic, shiny).\n",
            "   - There is a small green sphere (matte, non-metallic).\n",
            "   - There is a large cyan cube (matte, non-metallic).\n",
            "\n",
            "2. Now, apply the first instruction: \"Subtract all large yellow matte cubes.\"\n",
            "   - We look for objects that are: large, yellow, matte, and cubes.\n",
            "   - In the image, there is a large cube, but it is cyan (not yellow).\n",
            "   - Therefore, there are no objects matching this description.\n",
            "   - So, we subtract 0 objects in this step.\n",
            "\n",
            "3. Next, apply the second instruction: \"Subtract all metal things.\"\n",
            "   - We look for objects that are metallic.\n",
            "   - The gold cylinder is metallic.\n",
            "   - The red sphere is metallic.\n",
            "   - The green sphere and the cyan cube are matte (non-metallic).\n",
            "   - So, we subtract the gold cylinder and the red sphere.\n",
            "   - That means we remove 2 objects.\n",
            "\n",
            "4. Count the remaining objects:\n",
            "   - Initially, there were 4 objects.\n",
            "   - We subtracted 0 from the first instruction.\n",
            "   - We subtracted 2 from the second instruction.\n",
            "   - So, 4 - 0 - 2 = 2 objects remain.\n",
            "   - The remaining objects are the green sphere and the cyan cube.\n",
            "\n",
            "Therefore, 2 objects are left.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>2.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Subtract all large purple spheres. Subtract all small gray things. How many objects are left?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "9 \n",
            "Response:<REASONING>\n",
            "First, I need to identify all the objects in the image and categorize them by size and color to determine which ones to subtract.\n",
            "\n",
            "Total objects visible:\n",
            "1. Small blue cube (left side)\n",
            "2. Small red sphere (left side, behind the blue cube)\n",
            "3. Large shiny blue sphere (left side, behind the small red sphere)\n",
            "4. Large metallic silver cube (center-left)\n",
            "5. Large cyan cube (center-back)\n",
            "6. Small gray cylinder (center, behind the silver cube)\n",
            "7. Large tan cylinder (center-right)\n",
            "8. Small red cylinder (center-right, in front of the tan cylinder)\n",
            "9. Large green metallic cube (right side)\n",
            "10. Large tan sphere (front-center)\n",
            "\n",
            "That's 10 objects in total.\n",
            "\n",
            "Now, apply the subtraction rules:\n",
            "\n",
            "1. \"Subtract all large purple spheres\"\n",
            "   - Looking at the objects, there are no purple spheres. The large spheres are blue and tan. So, we subtract 0 objects here.\n",
            "\n",
            "2. \"Subtract all small gray things\"\n",
            "   - There is one small gray cylinder (the one behind the silver cube). So, we subtract 1 object.\n",
            "\n",
            "Therefore, we start with 10 objects and subtract 1, leaving 9 objects.\n",
            "\n",
            "No other objects match the description for subtraction.\n",
            "\n",
            "Final count: 10 - 1 = 9.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>9.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the age gap between these two people in image?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "4 \n",
            "Response:<REASONING>\n",
            "To estimate the age gap between the two people in the image, I would need to make visual assessments based on their physical appearance — such as facial features, skin texture, hair condition, and overall vitality — and compare them to typical age-related changes. Since there is no explicit date of birth or age information provided, I must rely on relative indicators.\n",
            "\n",
            "Observing the man: He appears to be in his late 50s to early 70s. His hair is thinning significantly on top, but he has some hair on the sides and back. His skin shows some wrinkles and signs of aging, but he appears to be in good health. His facial structure suggests maturity but not extreme old age.\n",
            "\n",
            "Observing the woman: She appears to be in her late 40s to early 60s. Her hair is curly and full, though possibly graying at the temples. She has a more youthful facial structure with less pronounced wrinkles than the man, and she is smiling brightly, suggesting vitality. Her skin tone appears more even and less wrinkled than the man’s.\n",
            "\n",
            "Comparing them: The woman appears to be younger than the man by approximately 10 to 15 years. The man’s more advanced signs of aging — particularly the balding and deeper wrinkles — suggest he is older than the woman. The woman’s full, curly hair and more youthful skin tone support her being younger.\n",
            "\n",
            "Therefore, I estimate the age gap to be around 12.5 years, as a reasonable average between the lower (10 years) and upper (15 years) estimate range, based on visual cues.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>12.5</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the sum of all the values in the ruling group?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "12 \n",
            "Response:<REASONING>\n",
            "To find the sum of all the values in the \"ruling\" group, I need to:\n",
            "1. Identify the \"ruling\" category on the y-axis.\n",
            "2. Read the value of the red bar (representing \"tin\") and the cyan bar (representing \"ace\") for the \"ruling\" group.\n",
            "3. Add these two values together.\n",
            "\n",
            "From the chart:\n",
            "- The red bar (\"tin\") for \"ruling\" extends to approximately 3.\n",
            "- The cyan bar (\"ace\") for \"ruling\" extends to approximately 9.\n",
            "\n",
            "Sum = 3 + 9 = 12.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>12.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>How many years have value less than 10%?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "5 \n",
            "Response:<REASONING>\n",
            "To determine how many years have a value less than 10%, I will examine the bar chart and list the percentage change for each year from 2010 to 2019:\n",
            "\n",
            "- 2010: 21.4% (≥10%)\n",
            "- 2011: 8.8% (<10%)\n",
            "- 2012: 16.9% (≥10%)\n",
            "- 2013: 12.1% (≥10%)\n",
            "- 2014: 5.3% (<10%)\n",
            "- 2015: 27.6% (≥10%)\n",
            "- 2016: 15.6% (≥10%)\n",
            "- 2017: 8.5% (<10%)\n",
            "- 2018: 2.4% (<10%)\n",
            "- 2019: -6.8% (<10%)\n",
            "\n",
            "Now, I count the years with values less than 10%: 2011, 2014, 2017, 2018, and 2019. That's a total of 5 years.\n",
            "\n",
            "Therefore, the answer is 5.\n",
            "</REASONING>\n",
            "<SOLUTION>5.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>如图，AC，BD是菱形ABCD的对角线，BH⊥AD于点H，若AC＝4，BD＝3，则BH的长为（）. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "2.4 \n",
            "Response:<REASONING>\n",
            "We are given a rhombus ABCD with diagonals AC = 4 and BD = 3. The diagonals of a rhombus bisect each other at right angles. Therefore, they intersect at point O (midpoint of both diagonals), forming four right triangles.\n",
            "\n",
            "Since AC = 4, AO = OC = 2.\n",
            "Since BD = 3, BO = OD = 1.5.\n",
            "\n",
            "The area of the rhombus can be calculated in two ways:\n",
            "1. Using the diagonals: Area = (d1 * d2) / 2 = (4 * 3) / 2 = 6.\n",
            "2. Using base and height: Area = base * height = AD * BH.\n",
            "\n",
            "We need to find the length of side AD first. In triangle AOD (a right triangle), we have:\n",
            "AO = 2, OD = 1.5.\n",
            "Using Pythagoras: AD = √(AO² + OD²) = √(2² + 1.5²) = √(4 + 2.25) = √6.25 = 2.5.\n",
            "\n",
            "Now, since Area = AD * BH, we have:\n",
            "6 = 2.5 * BH\n",
            "Therefore, BH = 6 / 2.5 = 2.4.\n",
            "\n",
            "Alternatively, we can use the formula for area = base * height, and since we have the area from the diagonals, we can directly compute BH as the height corresponding to base AD.\n",
            "\n",
            "</REASONING>\n",
            "\n",
            "<SOLUTION>\n",
            "2.4\n",
            "</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the y coordinate of the center of mass of the isosceles right triangle of uniform areal density shown in Figure 9-C?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "0.24 \n",
            "Response:<REASONING>\n",
            "To find the y-coordinate of the center of mass (centroid) of the isosceles right triangle shown in Figure 9-C, we can use the formula for the centroid of a triangle with respect to its base and height.\n",
            "\n",
            "The triangle is symmetric about the y-axis, so the x-coordinate of the center of mass is 0. We only need to find the y-coordinate.\n",
            "\n",
            "For a triangle with base b and height h, the centroid (center of mass) lies at a distance of h/3 from the base, measured along the median from the base to the opposite vertex.\n",
            "\n",
            "In this figure:\n",
            "- The triangle is isosceles right-angled, with the right angle at the top vertex (on the y-axis).\n",
            "- The two equal sides are labeled 'a'. Since it's a right triangle with legs equal to 'a', the height from the right angle to the hypotenuse is not 'a', but we need to be careful.\n",
            "\n",
            "Actually, let's reorient: the triangle has vertices at (0,0), (x,0), and (0,y). But from the figure, the two equal sides are labeled 'a', and the triangle is symmetric about the y-axis. So, the two legs are equal, and the right angle is at the top vertex (0, h), and the base is along the x-axis from (-b, 0) to (b, 0).\n",
            "\n",
            "Wait, let's read the figure: the triangle has two sides labeled 'a', both going down from the top vertex. So, the two equal sides are the legs of the right triangle. Therefore, the right angle is at the top vertex, and the base is the horizontal side.\n",
            "\n",
            "So, the triangle has:\n",
            "- Vertex at top: (0, h)\n",
            "- Vertex at bottom-left: (-b, 0)\n",
            "- Vertex at bottom-right: (b, 0)\n",
            "\n",
            "The two equal sides are from (0,h) to (-b,0) and from (0,h) to (b,0). The length of each is 'a'. So, distance from (0,h) to (-b,0) is sqrt((0 - (-b))^2 + (h - 0)^2) = sqrt(b^2 + h^2) = a.\n",
            "\n",
            "But since it's a right triangle with right angle at the top, the two legs are vertical and horizontal? No, that would make the right angle at the bottom. Let me think.\n",
            "\n",
            "Actually, in an isosceles right triangle, the right angle is between the two equal sides. So, if the two equal sides are labeled 'a', then the right angle is at the vertex where they meet. Looking at the figure, the top vertex is the right angle, and the two sides going down are both 'a'. So, the base is the horizontal side connecting the two bottom vertices.\n",
            "\n",
            "The length of the base is the distance between the two bottom vertices. Since the triangle is symmetric about the y-axis, the bottom vertices are at (-b, 0) and (b, 0), so the base length is 2b.\n",
            "\n",
            "The height is the perpendicular distance from the top vertex to the base. Since the top vertex is at (0, h), and the base is at y=0, the height is h.\n",
            "\n",
            "The two equal sides are from (0,h) to (-b,0) and from (0,h) to (b,0). The length of each is sqrt(b^2 + h^2) = a.\n",
            "\n",
            "But in an isosceles right triangle, the two legs are equal, and the hypotenuse is the base. Wait, no. In a right triangle, the hypotenuse is the side opposite the right angle. So, if the right angle is at the top, then the two legs are the two sides going down, and the base is the hypotenuse.\n",
            "\n",
            "So, the two legs are of length 'a', and the hypotenuse is the base.\n",
            "\n",
            "The length of the hypotenuse is sqrt(a^2 + a^2) = a*sqrt(2).\n",
            "\n",
            "The height from the right angle (top vertex) to the hypotenuse (base) is not h, but the altitude to the hypotenuse.\n",
            "\n",
            "In a right triangle, the altitude to the hypotenuse can be found using area.\n",
            "\n",
            "Area = (1/2) * leg1 * leg2 = (1/2) * a * a = a^2/2.\n",
            "\n",
            "Also, Area = (1/2) * hypotenuse * altitude to hypotenuse = (1/2) * (a*sqrt(2)) * h_hypotenuse.\n",
            "\n",
            "So, a^2/2 = (1/2) * a*sqrt(2) * h_hypotenuse\n",
            "\n",
            "=> a^2 = a*sqrt(2) * h_hypotenuse\n",
            "\n",
            "=> h_hypotenuse = a /\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the age gap between these two people in image?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "19 \n",
            "Response:<REASONING>\n",
            "To determine the age gap between the two individuals in the image, I need to identify who they are and their birth years, then calculate the difference.\n",
            "\n",
            "1. The woman on the left is Coretta Scott King. She was born on April 27, 1927, and died on January 30, 2006.\n",
            "2. The woman on the right is Hillary Clinton. She was born on October 20, 1947.\n",
            "\n",
            "Now, calculate the age difference:\n",
            "- Hillary Clinton's birth year: 1947\n",
            "- Coretta Scott King's birth year: 1927\n",
            "- Age gap = 1947 - 1927 = 20 years.\n",
            "\n",
            "Since Coretta Scott King was born earlier, she is older than Hillary Clinton by 20 years.\n",
            "\n",
            "Therefore, the age gap is 20 years.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>20.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Can you find the missing term?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "10 \n",
            "Response:<REASONING>\n",
            "To solve this, I need to find a pattern or relationship between the numbers in each row.\n",
            "\n",
            "Looking at the first row: 6, 9, 15\n",
            "- 6 + 9 = 15 → This suggests that the third number is the sum of the first two.\n",
            "\n",
            "Second row: 8, 12, 20\n",
            "- 8 + 12 = 20 → Again, the third number is the sum of the first two.\n",
            "\n",
            "So, if the pattern holds, for the third row: 4, 6, ?\n",
            "- 4 + 6 = 10\n",
            "\n",
            "Therefore, the missing term should be 10.\n",
            "\n",
            "I should also check if there are other possible patterns, such as multiplication or other operations, but the addition pattern is consistent across the first two rows and is the most straightforward.\n",
            "\n",
            "Let me verify: \n",
            "- Row 1: 6 + 9 = 15 ✓\n",
            "- Row 2: 8 + 12 = 20 ✓\n",
            "- Row 3: 4 + 6 = 10 ✓\n",
            "\n",
            "No other pattern seems to fit as cleanly. The pattern is clear: each row's third term is the sum of the first two terms.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>10.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the fraction of females facing the camera?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "0.8 \n",
            "Response:<REASONING>\n",
            "To solve this, I need to:\n",
            "1. Count the total number of people in the image.\n",
            "2. Count how many of them are females facing the camera.\n",
            "3. Calculate the fraction = (number of females facing camera) / (total number of people).\n",
            "\n",
            "Looking at the image:\n",
            "- There are 5 people standing together in the foreground.\n",
            "- All 5 are clearly female.\n",
            "- All 5 are facing the camera and smiling or looking at the camera.\n",
            "- So, all 5 are females facing the camera.\n",
            "\n",
            "Therefore, the fraction is 5/5 = 1.0.\n",
            "\n",
            "</REASONING>\n",
            "<SOLUTION>1.0</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>What is the size of the semicircle rounded to 2 decimal places?. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "14.14 \n",
            "Response:<REASONING>\n",
            "The given function is f(x) = √(9 - (x - 3)²). This represents the upper half of a circle.\n",
            "\n",
            "We can rewrite the equation in standard form of a circle:\n",
            "y = √(9 - (x - 3)²)\n",
            "Squaring both sides: y² = 9 - (x - 3)²\n",
            "Rearranging: (x - 3)² + y² = 9\n",
            "\n",
            "This is the equation of a circle with center at (3, 0) and radius squared = 9, so radius r = √9 = 3.\n",
            "\n",
            "The graph shows only the upper half of this circle (since y = √(...)), which is a semicircle.\n",
            "\n",
            "The size of the semicircle typically refers to its area.\n",
            "\n",
            "Area of a full circle = π * r² = π * 3² = 9π\n",
            "Area of a semicircle = (1/2) * 9π = 4.5π\n",
            "\n",
            "Now, we calculate the numerical value:\n",
            "4.5 * π ≈ 4.5 * 3.1415926535 ≈ 14.137167\n",
            "\n",
            "Rounded to 2 decimal places: 14.14\n",
            "\n",
            "Note: The graph shows the semicircle from x=0 to x=6, which is consistent with center at x=3 and radius 3. The shaded region is the right half of the semicircle, but the question asks for \"the size of the semicircle\", which is the area of the entire upper half-circle.\n",
            "\n",
            "</REASONING>\n",
            "\n",
            "<SOLUTION>\n",
            "14.14\n",
            "</SOLUTION>\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>At time $t=0$ a tank contains $Q_0 \\mathrm{lb}$ of salt dissolved in 100 gal of water; see Figure 2.3.1. Assume that water containing $\\frac{1}{4} \\mathrm{lb}$ of salt/gal is entering the tank at a rate of $r \\mathrm{gal} / \\mathrm{min}$ and that the well-stirred mixture is draining from the tank at the same rate. Set up the initial value problem that describes this flow process. By finding the amount of salt $Q(t)$ in the tank at any time, and the limiting amount $Q_L$ that is present after a very long time, if $r=3$ and $Q_0=2 Q_L$, find the time $T$ after which the salt level is within $2 \\%$ of $Q_L$. . Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "130.4 \n",
            "Response:<REASONING>\n",
            "This is a classic mixing problem in differential equations. The key idea is to model the rate of change of salt in the tank.\n",
            "\n",
            "Let Q(t) be the amount of salt (in lb) in the tank at time t (in min).\n",
            "\n",
            "The rate of change of salt in the tank is given by:\n",
            "dQ/dt = (rate in) - (rate out)\n",
            "\n",
            "Rate in: Water enters at r gal/min, and each gallon contains 1/4 lb of salt. So, rate in = r * (1/4) = r/4 lb/min.\n",
            "\n",
            "Rate out: The mixture drains at r gal/min. Since the tank is well-stirred, the concentration of salt in the tank at time t is Q(t)/100 lb/gal (because the volume of water is constant at 100 gal). So, rate out = r * (Q(t)/100) = rQ(t)/100 lb/min.\n",
            "\n",
            "Therefore, the differential equation is:\n",
            "dQ/dt = r/4 - (r/100) * Q(t)\n",
            "\n",
            "This is a linear first-order differential equation. We can write it as:\n",
            "dQ/dt + (r/100) * Q = r/4\n",
            "\n",
            "We can solve this using an integrating factor. The integrating factor is:\n",
            "μ(t) = exp(∫ (r/100) dt) = exp(rt/100)\n",
            "\n",
            "Multiplying both sides by μ(t):\n",
            "exp(rt/100) * dQ/dt + (r/100) * exp(rt/100) * Q = (r/4) * exp(rt/100)\n",
            "\n",
            "The left side is the derivative of (Q * exp(rt/100)):\n",
            "d/dt [Q * exp(rt/100)] = (r/4) * exp(rt/100)\n",
            "\n",
            "Integrate both sides:\n",
            "Q * exp(rt/100) = ∫ (r/4) * exp(rt/100) dt + C\n",
            "\n",
            "Let’s compute the integral:\n",
            "∫ (r/4) * exp(rt/100) dt = (r/4) * (100/r) * exp(rt/100) + C = (100/4) * exp(rt/100) + C = 25 * exp(rt/100) + C\n",
            "\n",
            "So,\n",
            "Q * exp(rt/100) = 25 * exp(rt/100) + C\n",
            "\n",
            "Therefore,\n",
            "Q(t) = 25 + C * exp(-rt/100)\n",
            "\n",
            "Now apply the initial condition: at t=0, Q(0) = Q0.\n",
            "Q0 = 25 + C * exp(0) => Q0 = 25 + C => C = Q0 - 25\n",
            "\n",
            "So the solution is:\n",
            "Q(t) = 25 + (Q0 - 25) * exp(-rt/100)\n",
            "\n",
            "The limiting amount Q_L is the amount of salt as t → ∞. Since exp(-rt/100) → 0 as t → ∞, we have:\n",
            "Q_L = 25 lb.\n",
            "\n",
            "We are given that r = 3 and Q0 = 2 * Q_L = 2 * 25 = 50 lb.\n",
            "\n",
            "So, Q(t) = 25 + (50 - 25) * exp(-3t/100) = 25 + 25 * exp(-3t/100)\n",
            "\n",
            "We want to find the time T after which the salt level is within 2% of Q_L. That is, we want Q(t) to be within 2% of 25, so:\n",
            "|Q(t) - 25| < 0.02 * 25 = 0.5\n",
            "\n",
            "Since Q(t) is increasing from 25 (because Q0=50 > 25) towards 25, we want:\n",
            "Q(t) - 25 < 0.5\n",
            "=> 25 * exp(-3t/100) < 0.5\n",
            "=> exp(-3t/100) < 0.5 / 25 = 0.02\n",
            "=> -3t/100 < ln(0.02)\n",
            "=> t > - (100/3) * ln(0.02)\n",
            "\n",
            "Let’s compute this value.\n",
            "\n",
            "ln(0.02) = ln(2/100) = ln(2) - ln(100) ≈ 0.693147 - 4.605170 = -3.912023\n",
            "\n",
            "So,\n",
            "t > - (100/\n",
            "-------------------- Question:\n",
            "<|im_start|>user\n",
            "<|vision_start|><|image_pad|><|vision_end|>Find x. Round to the nearest tenth, if necessary.. Also first provide your reasoning or working out on how you would go about solving the question between <REASONING> and </REASONING> and then your final answer between <SOLUTION> and (put a single float here) </SOLUTION><|im_end|>\n",
            "<|im_start|>assistant\n",
            " \n",
            "Answer:\n",
            "3 \n",
            "Response:<REASONING>\n",
            "This problem involves a circle with a tangent and a secant intersecting at a point outside the circle. The tangent segment is labeled 9, and the secant segment is divided into two parts: the external part is labeled x, and the entire secant from the external point to the far intersection point is labeled 2x. The length of the part of the secant inside the circle is labeled 12.25.\n",
            "\n",
            "According to the Power of a Point Theorem for a tangent and a secant intersecting at a point outside a circle, the square of the length of the tangent segment is equal to the product of the lengths of the entire secant segment and its external part.\n",
            "\n",
            "So, the formula is:\n",
            "(tangent)^2 = (external part of secant) × (entire secant)\n",
            "\n",
            "In this case:\n",
            "9^2 = x × (x + 12.25)\n",
            "\n",
            "Wait, let me check the diagram. The secant has two parts: from the external point to the first intersection point is labeled x, and from the first intersection point to the second intersection point is labeled 12.25. So the entire secant from the external point to the second intersection point is x + 12.25.\n",
            "\n",
            "Therefore, the equation is:\n",
            "9^2 = x × (x + 12.25)\n",
            "\n",
            "Let me solve this equation:\n",
            "81 = x^2 + 12.25x\n",
            "\n",
            "Bring all terms to one side:\n",
            "x^2 + 12.25x - 81 = 0\n",
            "\n",
            "This is a quadratic equation. I will solve it using the quadratic formula:\n",
            "x = [-b ± sqrt(b^2 - 4ac)] / (2a)\n",
            "where a = 1, b = 12.25, c = -81.\n",
            "\n",
            "First, calculate the discriminant:\n",
            "D = (12.25)^2 - 4(1)(-81)\n",
            "D = 150.0625 + 324\n",
            "D = 474.0625\n",
            "\n",
            "Now, sqrt(474.0625) = 21.773... (since 21.773^2 ≈ 474.0625)\n",
            "\n",
            "So,\n",
            "x = [-12.25 ± 21.773] / 2\n",
            "\n",
            "We take the positive root since length cannot be negative:\n",
            "x = (-12.25 + 21.773) / 2\n",
            "x = 9.523 / 2\n",
            "x = 4.7615\n",
            "\n",
            "Rounding to the nearest tenth: 4.8\n",
            "\n",
            "Wait, let me verify the diagram. The problem says the secant is labeled \"2x\" and \"y\" is also there, but the diagram shows \"2x\" as the entire secant from the external point to the far point, and \"y\" is the segment from the external point to the first intersection point? That would make \"y\" = 2x? That doesn't make sense.\n",
            "\n",
            "Looking at the diagram again: There is a point outside the circle. From that point, a tangent goes down to the circle, labeled 9. From that same point, a secant goes through the circle, intersecting it at two points. The segment from the external point to the first intersection point is labeled \"x\". The segment from the first intersection point to the second intersection point is labeled \"12.25\". The entire secant from the external point to the second intersection point is labeled \"2x\".\n",
            "\n",
            "Wait, that can't be. If the entire secant is 2x, and the external part is x, then the internal part must be x as well. But the internal part is labeled 12.25. So x = 12.25? That contradicts the diagram.\n",
            "\n",
            "Let me re-read the problem. The diagram shows:\n",
            "- A tangent segment of length 9.\n",
            "- A secant segment labeled \"2x\" (from the external point to the far point on the circle).\n",
            "- The secant is divided into two parts: from external point to first intersection: labeled \"x\", and from first intersection to second intersection: labeled \"12.25\".\n",
            "- So the entire secant is x + 12.25.\n",
            "- The problem says the entire secant is \"2x\".\n",
            "\n",
            "So, x + 12.25 = 2x.\n",
            "\n",
            "That would mean 12.25 = x.\n",
            "\n",
            "But then the tangent is 9, and according to Power of a Point, 9^2 = x * (x + 12.25) = x * 2x = 2x^2.\n",
            "\n",
            "So 81 = 2x^2, x\n"
          ]
        },
        {
          "data": {
            "text/plain": [
              "TrainOutput(global_step=60, training_loss=0.010748568288023765, metrics={'train_runtime': 6405.6326, 'train_samples_per_second': 0.019, 'train_steps_per_second': 0.009, 'total_flos': 0.0, 'train_loss': 0.010748568288023765})"
            ]
          },
          "execution_count": 19,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "trainer = GRPOTrainer(\n",
        "    model = model,\n",
        "    args = training_args,\n",
        "    # Pass the processor to handle multimodal inputs\n",
        "    processing_class = tokenizer,\n",
        "    reward_funcs = [\n",
        "        formatting_reward_func,\n",
        "        correctness_reward_func,\n",
        "    ],\n",
        "    train_dataset = train_dataset,\n",
        ")\n",
        "\n",
        "trainer.train()"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "tlaUdxC_VHpz"
      },
      "source": [
        "<a name=\"Inference\"></a>\n",
        "### Inference\n",
        "\n"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "LzUvkjO6ffIs"
      },
      "source": [
        "Let's run the model! You can modify the instruction and input."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 27,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "qtcz_lpbVC92",
        "outputId": "7dab8bf6-22c3-4ec1-a4e6-2a2af91ce3fc"
      },
      "outputs": [
        {
          "name": "stdout",
          "output_type": "stream",
          "text": [
            "<REASONING>\n",
            "To find the magnitude of the average force on the driver during the collision, we need to apply the impulse-momentum theorem. The impulse (change in momentum) equals the average force multiplied by the time interval over which the force acts.\n",
            "\n",
            "The change in momentum is given by:\n",
            "Δp = m * Δv = m * (v_f - v_i)\n",
            "\n",
            "However, since the velocities are vectors (with directions), we must resolve them into components perpendicular and parallel to the wall, because the wall exerts a force only perpendicular to the wall (normal force), and the parallel component of velocity typically does not change in an ideal elastic collision (though here it's inelastic, the parallel component might still be unchanged if the wall is perfectly rigid and the collision is only normal).\n",
            "\n",
            "In this case, the car’s velocity has components parallel and perpendicular to the wall. The wall only exerts a force perpendicular to itself. Therefore, the component of velocity parallel to the wall should remain unchanged, and only the component perpendicular to the wall changes.\n",
            "\n",
            "Let’s define:\n",
            "- The wall is along the x-axis (as shown in the diagram, with the wall horizontal, and y-axis perpendicular to it).\n",
            "- The car’s initial velocity vector makes an angle of 30° with the wall, so the angle with the y-axis (perpendicular to wall) is 60°, but we can directly use the angle given with respect to the wall.\n",
            "\n",
            "Actually, looking at the diagram:\n",
            "- The initial velocity is at 30° from the wall (so 30° from the x-axis).\n",
            "- The final velocity is at 10° from the wall (so 10° from the x-axis).\n",
            "\n",
            "Therefore:\n",
            "- The component of velocity parallel to the wall (x-direction) is: v_x = v * cos(θ)\n",
            "- The component of velocity perpendicular to the wall (y-direction) is: v_y = v * sin(θ)\n",
            "\n",
            "The wall exerts a force only in the y-direction (perpendicular to the wall). Therefore, the change in momentum in the y-direction will be what matters for the average force, because the x-component doesn't change (no force in x-direction from the wall).\n",
            "\n",
            "So, let's compute the initial and final y-components of velocity:\n",
            "\n",
            "Initial velocity:\n",
            "v_i = 70 m/s at 30° from the wall (x-axis)\n",
            "So, initial y-component: v_{iy} = 70 * sin(30°) = 70 * 0.5 = 35 m/s (in the positive y-direction, since the car is approaching the wall from below, and the y-axis is upward)\n",
            "\n",
            "Final velocity:\n",
            "v_f = 50 m/s at 10° from the wall (x-axis)\n",
            "So, final y-component: v_{fy} = 50 * sin(10°) ≈ 50 * 0.173648 ≈ 8.6824 m/s (but note: after collision, the car is moving away from the wall? Or is it still approaching? The diagram shows the car moving away from the wall after collision, since it's shown going away from the wall. But if the car is colliding with the wall and then continues, the y-component should be negative after collision? Wait, let's think.\n",
            "\n",
            "Looking at the diagram:\n",
            "- The car is approaching the wall from below (the arrow points toward the wall from the lower left, at 30° to the x-axis). So, the y-component is downward, meaning negative if we take upward as positive y.\n",
            "- After collision, the car is moving away from the wall, to the right and slightly upward (10° from the x-axis). So the y-component is upward, positive.\n",
            "\n",
            "But the initial y-component is downward (toward the wall), so it should be negative. Let's define the y-axis as positive upward. Then:\n",
            "\n",
            "Initial y-component: v_{iy} = -70 * sin(30°) = -35 m/s\n",
            "Final y-component: v_{fy} = +50 * sin(10°) ≈ +8.6824 m/s\n",
            "\n",
            "Therefore, the change in y-component of velocity: Δv_y = v_{fy} - v_{iy} = 8.6824 - (-35) = 8.6824 + 35 = 43.6824 m/s\n",
            "\n",
            "Then, change in momentum in y-direction: Δp_y = m * Δv_y = 80 * 43.6824 = 3494.592 kg·m/s\n",
            "\n",
            "The time interval of collision: Δt = 14 ms = 0.014 s\n",
            "\n",
            "Average force in y-direction: F_avg = Δp_y / Δt = 3494.\n"
          ]
        }
      ],
      "source": [
        "image = train_dataset[165][\"image\"]\n",
        "prompt = train_dataset[165][\"prompt\"]\n",
        "\n",
        "inputs = tokenizer(\n",
        "    image,\n",
        "    prompt,\n",
        "    add_special_tokens = False,\n",
        "    return_tensors = \"pt\",\n",
        ").to(\"cuda\")\n",
        "\n",
        "from transformers import TextStreamer\n",
        "text_streamer = TextStreamer(tokenizer, skip_prompt = True)\n",
        "_ = model.generate(**inputs, streamer = text_streamer, max_new_tokens = 1024,\n",
        "                   use_cache = True, temperature = 1.0, min_p = 0.1)"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "Colxz9TAVMsi"
      },
      "source": [
        "<a name=\"Save\"></a>\n",
        "### Saving, loading finetuned models\n",
        "To save the final model as LoRA adapters, use Hugging Face’s `push_to_hub` for online saving, or `save_pretrained` for local storage.\n",
        "\n",
        "**[NOTE]** This ONLY saves the LoRA adapters, and not the full model. To save to 16bit or GGUF, scroll down!"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 21,
      "metadata": {
        "colab": {
          "base_uri": "https://localhost:8080/"
        },
        "id": "AL-BcuB1VLIv",
        "outputId": "905cc0c8-ebb9-4e2a-8b13-db77c054c222"
      },
      "outputs": [
        {
          "data": {
            "text/plain": [
              "[]"
            ]
          },
          "execution_count": 21,
          "metadata": {},
          "output_type": "execute_result"
        }
      ],
      "source": [
        "model.save_pretrained(\"grpo_lora\")  # Local saving\n",
        "tokenizer.save_pretrained(\"grpo_lora\")\n",
        "# model.push_to_hub(\"your_name/grpo_lora\", token = \"...\") # Online saving\n",
        "# processor.push_to_hub(\"your_name/grpo_lora\", token = \"...\") # Online saving"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "a4LMOBl8boGX"
      },
      "source": [
        "Verify LoRA is actually trained!"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 23,
      "metadata": {
        "id": "4SfdI-ERbpiw"
      },
      "outputs": [],
      "source": [
        "from safetensors import safe_open\n",
        "\n",
        "tensors = {}\n",
        "with safe_open(\"grpo_lora/adapter_model.safetensors\", framework = \"pt\") as f:\n",
        "    # Verify both A and B are non zero\n",
        "    for key in f.keys():\n",
        "        tensor = f.get_tensor(key)\n",
        "        n_zeros = (tensor == 0).sum() / tensor.numel()\n",
        "        assert(n_zeros.item() != tensor.numel())"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "-NUEmHFSYNTp"
      },
      "source": [
        "<a name=\"Save\"></a>\n",
        "### Saving to float16 for VLLM\n",
        "\n",
        "We also support saving to `float16` directly. Select `merged_16bit` for float16 or `merged_4bit` for int4. We also allow `lora` adapters as a fallback. Use `push_to_hub_merged` to upload to your Hugging Face account! You can go to https://huggingface.co/settings/tokens for your personal tokens."
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 24,
      "metadata": {
        "id": "07lMVV96vz39"
      },
      "outputs": [],
      "source": [
        "# Merge to 16bit\n",
        "if False: model.save_pretrained_merged(\"model\", tokenizer, save_method = \"merged_16bit\",)\n",
        "if False: model.push_to_hub_merged(\"hf/model\", tokenizer, save_method = \"merged_16bit\", token = \"\")\n",
        "\n",
        "# Merge to 4bit\n",
        "if False: model.save_pretrained_merged(\"model\", tokenizer, save_method = \"merged_4bit\",)\n",
        "if False: model.push_to_hub_merged(\"hf/model\", tokenizer, save_method = \"merged_4bit\", token = \"\")\n",
        "\n",
        "# Just LoRA adapters\n",
        "if False:\n",
        "    model.save_pretrained(\"model\")\n",
        "    tokenizer.save_pretrained(\"model\")\n",
        "if False:\n",
        "    model.push_to_hub(\"hf/model\", token = \"\")\n",
        "    tokenizer.push_to_hub(\"hf/model\", token = \"\")"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "52WMb3k_YPt8"
      },
      "source": [
        "### GGUF / llama.cpp Conversion\n",
        "To save to `GGUF` / `llama.cpp`, we support it natively now! We clone `llama.cpp` and we default save it to `q8_0`. We allow all methods like `q4_k_m`. Use `save_pretrained_gguf` for local saving and `push_to_hub_gguf` for uploading to HF.\n",
        "\n",
        "Some supported quant methods (full list on our [Wiki page](https://github.com/unslothai/unsloth/wiki#gguf-quantization-options)):\n",
        "* `q8_0` - Fast conversion. High resource use, but generally acceptable.\n",
        "* `q4_k_m` - Recommended. Uses Q6_K for half of the attention.wv and feed_forward.w2 tensors, else Q4_K.\n",
        "* `q5_k_m` - Recommended. Uses Q6_K for half of the attention.wv and feed_forward.w2 tensors, else Q5_K.\n",
        "\n",
        "[**NEW**] To finetune and auto export to Ollama, try our [Ollama notebook](https://colab.research.google.com/github/unslothai/notebooks/blob/main/nb/Llama3_(8B)-Ollama.ipynb)"
      ]
    },
    {
      "cell_type": "code",
      "execution_count": 25,
      "metadata": {
        "id": "QyEjW-WuYQIm"
      },
      "outputs": [],
      "source": [
        "# Save to 8bit Q8_0\n",
        "if False: model.save_pretrained_gguf(\"model\", tokenizer,)\n",
        "# Remember to go to https://huggingface.co/settings/tokens for a token!\n",
        "# And change hf to your username!\n",
        "if False: model.push_to_hub_gguf(\"hf/model\", tokenizer, token = \"\")\n",
        "\n",
        "# Save to 16bit GGUF\n",
        "if False: model.save_pretrained_gguf(\"model\", tokenizer, quantization_method = \"f16\")\n",
        "if False: model.push_to_hub_gguf(\"hf/model\", tokenizer, quantization_method = \"f16\", token = \"\")\n",
        "\n",
        "# Save to q4_k_m GGUF\n",
        "if False: model.save_pretrained_gguf(\"model\", tokenizer, quantization_method = \"q4_k_m\")\n",
        "if False: model.push_to_hub_gguf(\"hf/model\", tokenizer, quantization_method = \"q4_k_m\", token = \"\")\n",
        "\n",
        "# Save to multiple GGUF options - much faster if you want multiple!\n",
        "if False:\n",
        "    model.push_to_hub_gguf(\n",
        "        \"hf/model\", # Change hf to your username!\n",
        "        tokenizer,\n",
        "        quantization_method = [\"q4_k_m\", \"q8_0\", \"q5_k_m\",],\n",
        "        token = \"\",\n",
        "    )"
      ]
    },
    {
      "cell_type": "markdown",
      "metadata": {
        "id": "dbm6rx-dbt5Z"
      },
      "source": [
        "Special Credits to [GAD-Cell](https://github.com/GAD-cell) for helping Unsloth create this notebook and bringing VLM GRPO into Unsloth!"
      ]
    }
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